Variance of $XY(1-Y)$ in terms of the means and variances $X$ and $Y$












2














Consider two independent random variables X and Y. X has some distribution with mean $mu_X$ and variance $sigma^2_X$. Y has some distribution with mean $mu_Y$ and variance $sigma^2_Y$. I want to take the variance of the quantity $XY(1-Y)$. Is the correct answer:
$$sigma^2_{XY(1-Y)} = sigma^2_X[Y(1-Y)]^2+X^2[sigma^2_Y(1-sigma^2_Y)]?$$
I just did some kind of weird product rule but with variances and I'm not sure that's at all legit.



The reason I want to know the variance of this quantity is that it is a measure of the real noise in an experimental system I am working with. I am an experimental physicist and my math skills are weak at best. You might recognize the form $XY(1-Y)$ from the binomial distribution: $Np(1-p)$. My experiment does not follow a binomial distribution exactly, it follows a Poisson binomial distribution but with varying $N$ and $p$ for each shot of the experiment. So I need to know the variance of the quantity $XY(1-Y)$ when I know the variances and expectation values of $X$ and $Y$.



I'm trying to work this particular detail out to answer this question:
Repeated binomial processes with different probabilities and number of trials .






probability-theory variance







share|cite|improve this question




















  • 1




    Comments are not for extended discussion; this conversation has been moved to chat.
    – quid♦
    Dec 24 '18 at 14:55






  • 1




    The formula $$sigma^2_{XY(1-Y)} = sigma^2_X[Y(1-Y)]^2+X^2[sigma^2_Y(1-sigma^2_Y)]$$ cannot be correct since $sigma^2_{XY(1-Y)}$, $sigma^2_X$ and $sigma^2_Y(1-sigma^2_Y)$ are numbers while $[Y(1-Y)]^2$ and $X^2$ are random variables.
    – Did
    Dec 25 '18 at 18:08
















2














Consider two independent random variables X and Y. X has some distribution with mean $mu_X$ and variance $sigma^2_X$. Y has some distribution with mean $mu_Y$ and variance $sigma^2_Y$. I want to take the variance of the quantity $XY(1-Y)$. Is the correct answer:
$$sigma^2_{XY(1-Y)} = sigma^2_X[Y(1-Y)]^2+X^2[sigma^2_Y(1-sigma^2_Y)]?$$
I just did some kind of weird product rule but with variances and I'm not sure that's at all legit.



The reason I want to know the variance of this quantity is that it is a measure of the real noise in an experimental system I am working with. I am an experimental physicist and my math skills are weak at best. You might recognize the form $XY(1-Y)$ from the binomial distribution: $Np(1-p)$. My experiment does not follow a binomial distribution exactly, it follows a Poisson binomial distribution but with varying $N$ and $p$ for each shot of the experiment. So I need to know the variance of the quantity $XY(1-Y)$ when I know the variances and expectation values of $X$ and $Y$.



I'm trying to work this particular detail out to answer this question:
Repeated binomial processes with different probabilities and number of trials .






probability-theory variance







share|cite|improve this question




















  • 1




    Comments are not for extended discussion; this conversation has been moved to chat.
    – quid♦
    Dec 24 '18 at 14:55






  • 1




    The formula $$sigma^2_{XY(1-Y)} = sigma^2_X[Y(1-Y)]^2+X^2[sigma^2_Y(1-sigma^2_Y)]$$ cannot be correct since $sigma^2_{XY(1-Y)}$, $sigma^2_X$ and $sigma^2_Y(1-sigma^2_Y)$ are numbers while $[Y(1-Y)]^2$ and $X^2$ are random variables.
    – Did
    Dec 25 '18 at 18:08














2












2








2







Consider two independent random variables X and Y. X has some distribution with mean $mu_X$ and variance $sigma^2_X$. Y has some distribution with mean $mu_Y$ and variance $sigma^2_Y$. I want to take the variance of the quantity $XY(1-Y)$. Is the correct answer:
$$sigma^2_{XY(1-Y)} = sigma^2_X[Y(1-Y)]^2+X^2[sigma^2_Y(1-sigma^2_Y)]?$$
I just did some kind of weird product rule but with variances and I'm not sure that's at all legit.



The reason I want to know the variance of this quantity is that it is a measure of the real noise in an experimental system I am working with. I am an experimental physicist and my math skills are weak at best. You might recognize the form $XY(1-Y)$ from the binomial distribution: $Np(1-p)$. My experiment does not follow a binomial distribution exactly, it follows a Poisson binomial distribution but with varying $N$ and $p$ for each shot of the experiment. So I need to know the variance of the quantity $XY(1-Y)$ when I know the variances and expectation values of $X$ and $Y$.



I'm trying to work this particular detail out to answer this question:
Repeated binomial processes with different probabilities and number of trials .






probability-theory variance







share|cite|improve this question















Consider two independent random variables X and Y. X has some distribution with mean $mu_X$ and variance $sigma^2_X$. Y has some distribution with mean $mu_Y$ and variance $sigma^2_Y$. I want to take the variance of the quantity $XY(1-Y)$. Is the correct answer:
$$sigma^2_{XY(1-Y)} = sigma^2_X[Y(1-Y)]^2+X^2[sigma^2_Y(1-sigma^2_Y)]?$$
I just did some kind of weird product rule but with variances and I'm not sure that's at all legit.



The reason I want to know the variance of this quantity is that it is a measure of the real noise in an experimental system I am working with. I am an experimental physicist and my math skills are weak at best. You might recognize the form $XY(1-Y)$ from the binomial distribution: $Np(1-p)$. My experiment does not follow a binomial distribution exactly, it follows a Poisson binomial distribution but with varying $N$ and $p$ for each shot of the experiment. So I need to know the variance of the quantity $XY(1-Y)$ when I know the variances and expectation values of $X$ and $Y$.



I'm trying to work this particular detail out to answer this question:
Repeated binomial processes with different probabilities and number of trials .






probability-theory variance




probability-theory variance






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 18:09









Did

246k23221455




246k23221455










asked Dec 23 '18 at 5:42









SabrinaChoice

246




246








  • 1




    Comments are not for extended discussion; this conversation has been moved to chat.
    – quid♦
    Dec 24 '18 at 14:55






  • 1




    The formula $$sigma^2_{XY(1-Y)} = sigma^2_X[Y(1-Y)]^2+X^2[sigma^2_Y(1-sigma^2_Y)]$$ cannot be correct since $sigma^2_{XY(1-Y)}$, $sigma^2_X$ and $sigma^2_Y(1-sigma^2_Y)$ are numbers while $[Y(1-Y)]^2$ and $X^2$ are random variables.
    – Did
    Dec 25 '18 at 18:08














  • 1




    Comments are not for extended discussion; this conversation has been moved to chat.
    – quid♦
    Dec 24 '18 at 14:55






  • 1




    The formula $$sigma^2_{XY(1-Y)} = sigma^2_X[Y(1-Y)]^2+X^2[sigma^2_Y(1-sigma^2_Y)]$$ cannot be correct since $sigma^2_{XY(1-Y)}$, $sigma^2_X$ and $sigma^2_Y(1-sigma^2_Y)$ are numbers while $[Y(1-Y)]^2$ and $X^2$ are random variables.
    – Did
    Dec 25 '18 at 18:08








1




1




Comments are not for extended discussion; this conversation has been moved to chat.
– quid♦
Dec 24 '18 at 14:55




Comments are not for extended discussion; this conversation has been moved to chat.
– quid♦
Dec 24 '18 at 14:55




1




1




The formula $$sigma^2_{XY(1-Y)} = sigma^2_X[Y(1-Y)]^2+X^2[sigma^2_Y(1-sigma^2_Y)]$$ cannot be correct since $sigma^2_{XY(1-Y)}$, $sigma^2_X$ and $sigma^2_Y(1-sigma^2_Y)$ are numbers while $[Y(1-Y)]^2$ and $X^2$ are random variables.
– Did
Dec 25 '18 at 18:08




The formula $$sigma^2_{XY(1-Y)} = sigma^2_X[Y(1-Y)]^2+X^2[sigma^2_Y(1-sigma^2_Y)]$$ cannot be correct since $sigma^2_{XY(1-Y)}$, $sigma^2_X$ and $sigma^2_Y(1-sigma^2_Y)$ are numbers while $[Y(1-Y)]^2$ and $X^2$ are random variables.
– Did
Dec 25 '18 at 18:08










1 Answer
1






active

oldest

votes


















2














$begin{aligned}mathsf{Var}XYleft(1-Yright) & =mathsf{Covar}left(XYleft(1-Yright),XYleft(1-Yright)right)\
& =mathsf{Covar}left(XY-XY^{2},XY-XY^{2}right)\
& =mathsf{Var}left(XYright)-2mathsf{Covar}left(XY^{2},XYright)+mathsf{Var}left(XY^{2}right)\
& =left[mathbb{E}X^{2}Y^{2}-left(mathbb{E}XYright)^{2}right]-2left[mathbb{E}X^{2}Y^{3}-mathbb{E}XY^{2}mathbb{E}XYright]+left[mathbb{E}X^{2}Y^{4}-left(mathbb{E}XY^{2}right)^{2}right]\
& =left[mathbb{E}X^{2}mathbb{E}Y^{2}-left(mathbb{E}Xmathbb{E}Yright)^{2}right]-2left[mathbb{E}X^{2}mathbb{E}Y^{3}-mathbb{E}Xmathbb{E}Y^{2}mathbb{E}Xmathbb{E}Yright]+left[mathbb{E}X^{2}mathbb{E}Y^{4}-left(mathbb{E}Xmathbb{E}Y^{2}right)^{2}right]\
& =left[mathbb{E}X^{2}mathbb{E}Y^{2}-mu_{X}^{2}mu_{Y}^{2}right]-2left[mathbb{E}X^{2}mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}mathbb{E}Y^{2}right]+left[mathbb{E}X^{2}mathbb{E}Y^{4}-mu_{X}^{2}left(mathbb{E}Y^{2}right)^{2}right]\
& =left[left(sigma_{X}^{2}+mu_{X}^{2}right)left(sigma_{Y}^{2}+mu_{Y}^{2}right)-mu_{X}^{2}mu_{Y}^{2}right]-2left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}left(sigma_{Y}^{2}+mu_{Y}^{2}right)right]+left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{4}-mu_{X}^{2}left(sigma_{Y}^{2}+mu_{Y}^{2}right)^{2}right]\
& =left[sigma_{X}^{2}sigma_{Y}^{2}+sigma_{X}^{2}mu_{Y}^{2}+sigma_{Y}^{2}mu_{X}^{2}right]-2left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}left(sigma_{Y}^{2}+mu_{Y}^{2}right)right]+left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{4}-mu_{X}^{2}left(sigma_{Y}^{2}+mu_{Y}^{2}right)^{2}right]
end{aligned}
$



Especially the $5$-th equality rests on independence of $X$ and $Y$.



We cannot get any further since for the expressions $mathbb EY^3$ and $mathbb EY^4$ more information is needed concerning the distribution of $Y$.




edit:



Special case: $Ysimmathsf{Normal}left(mu_{Y},sigma_{Y}^{2}right)$.



Then $Y=mu_{Y}+sigma_{Y}U$ where $Usimmathsf{Normal}left(0,1right)$
and based on the knowledge that $mathbb{E}U^{3}=0$ and $mathbb{E}U^{4}=3$
we find:




  • $mathbb{E}Y^{3}=mathbb{E}left(mu_{Y}^{3}+3mu_{Y}^{2}sigma_{Y}U+3mu_{Y}sigma_{Y}^{2}U^{2}+sigma_{Y}^{3}U^{3}right)=mu_{Y}^{3}+3mu_{Y}sigma_{Y}^{2}$

  • $mathbb{E}Y^{4}=mathbb{E}left(mu_{Y}^{4}+4mu_{Y}^{3}sigma_{Y}U+6mu_{Y}^{2}sigma_{Y}^{2}U^{2}+4mu_{Y}sigma_{Y}^{3}U^{3}+sigma_{Y}^{4}U^{4}right)=mu_{Y}^{4}+6mu_{Y}^{2}sigma_{Y}^{2}+3sigma_{Y}^{4}$


Also have a look here for that.



Substituting we find the following expression for $mathsf{Var}XYleft(1-Yright)$:



$$left[sigma_{X}^{2}sigma_{Y}^{2}+sigma_{X}^{2}mu_{Y}^{2}+sigma_{Y}^{2}mu_{X}^{2}right]-2left[sigma_{X}^{2}mu_{Y}^{3}+3sigma_{X}^{2}sigma_{Y}^{2}mu_{Y}+2sigma_{Y}^{2}mu_{X}^{2}mu_{Y}right]+left[sigma_{X}^{2}mu_{Y}^{4}+6sigma_{X}^{2}sigma_{Y}^{2}mu_{Y}^{2}+3sigma_{X}^{2}sigma_{Y}^{4}+4sigma_{Y}^{2}mu_{X}^{2}mu_{Y}^{2}+2sigma_{Y}^{4}mu_{X}^{2}right]$$



I hope I did not make any mistakes (check me on it).






share|cite|improve this answer























  • This is wonderful, thank you! I wouldn't have thought of taking the covariance first. I'm not actually sure about the exact distributions of X and Y (they will be empirically determined). They'll probably end up Gaussian. Once I know them, then I would just have to do the integrals. Thanks so much, this is really helpful.
    – SabrinaChoice
    Dec 24 '18 at 14:47








  • 1




    @SabrinaChoice: In general, you can try to look at the moments of any sufficiently nice distribution if they exist, and each moment is a separate 'degree of freedom'. In fact, if all finite moments exist, then they uniquely characterize the distribution (if it has a continuous CDF). So we can also compare distributions by comparing their moments.
    – user21820
    Dec 24 '18 at 15:02






  • 1




    @SabrinaChoice I let you know that I found a mistake (and repaired). $mathbb EX^2mathbb EX^2$ was accidently replaced by $sigma_X^2+mu_X^2$, but had to be replaced by $(sigma_X^2+mu_X^2)(sigma_Y^2+mu_Y^2)$ of course.
    – drhab
    Dec 25 '18 at 16:44








  • 1




    In my former comment $mathbb EX^2mathbb EX^2$ should be $mathbb EX^2mathbb EY^2$. Also I edited for the special case that $Y$ takes normal distribution.
    – drhab
    Dec 26 '18 at 8:15












  • @drhab I finally had the chance to go through this carefully and check for errors. I couldn't find any. Thank you very much, I really appreciate it. My math skills are weak, but I feel like this particular example will make me much better equipped to solve a problem like this on my own next time I encounter one.
    – SabrinaChoice
    13 hours ago













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050093%2fvariance-of-xy1-y-in-terms-of-the-means-and-variances-x-and-y%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














$begin{aligned}mathsf{Var}XYleft(1-Yright) & =mathsf{Covar}left(XYleft(1-Yright),XYleft(1-Yright)right)\
& =mathsf{Covar}left(XY-XY^{2},XY-XY^{2}right)\
& =mathsf{Var}left(XYright)-2mathsf{Covar}left(XY^{2},XYright)+mathsf{Var}left(XY^{2}right)\
& =left[mathbb{E}X^{2}Y^{2}-left(mathbb{E}XYright)^{2}right]-2left[mathbb{E}X^{2}Y^{3}-mathbb{E}XY^{2}mathbb{E}XYright]+left[mathbb{E}X^{2}Y^{4}-left(mathbb{E}XY^{2}right)^{2}right]\
& =left[mathbb{E}X^{2}mathbb{E}Y^{2}-left(mathbb{E}Xmathbb{E}Yright)^{2}right]-2left[mathbb{E}X^{2}mathbb{E}Y^{3}-mathbb{E}Xmathbb{E}Y^{2}mathbb{E}Xmathbb{E}Yright]+left[mathbb{E}X^{2}mathbb{E}Y^{4}-left(mathbb{E}Xmathbb{E}Y^{2}right)^{2}right]\
& =left[mathbb{E}X^{2}mathbb{E}Y^{2}-mu_{X}^{2}mu_{Y}^{2}right]-2left[mathbb{E}X^{2}mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}mathbb{E}Y^{2}right]+left[mathbb{E}X^{2}mathbb{E}Y^{4}-mu_{X}^{2}left(mathbb{E}Y^{2}right)^{2}right]\
& =left[left(sigma_{X}^{2}+mu_{X}^{2}right)left(sigma_{Y}^{2}+mu_{Y}^{2}right)-mu_{X}^{2}mu_{Y}^{2}right]-2left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}left(sigma_{Y}^{2}+mu_{Y}^{2}right)right]+left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{4}-mu_{X}^{2}left(sigma_{Y}^{2}+mu_{Y}^{2}right)^{2}right]\
& =left[sigma_{X}^{2}sigma_{Y}^{2}+sigma_{X}^{2}mu_{Y}^{2}+sigma_{Y}^{2}mu_{X}^{2}right]-2left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}left(sigma_{Y}^{2}+mu_{Y}^{2}right)right]+left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{4}-mu_{X}^{2}left(sigma_{Y}^{2}+mu_{Y}^{2}right)^{2}right]
end{aligned}
$



Especially the $5$-th equality rests on independence of $X$ and $Y$.



We cannot get any further since for the expressions $mathbb EY^3$ and $mathbb EY^4$ more information is needed concerning the distribution of $Y$.




edit:



Special case: $Ysimmathsf{Normal}left(mu_{Y},sigma_{Y}^{2}right)$.



Then $Y=mu_{Y}+sigma_{Y}U$ where $Usimmathsf{Normal}left(0,1right)$
and based on the knowledge that $mathbb{E}U^{3}=0$ and $mathbb{E}U^{4}=3$
we find:




  • $mathbb{E}Y^{3}=mathbb{E}left(mu_{Y}^{3}+3mu_{Y}^{2}sigma_{Y}U+3mu_{Y}sigma_{Y}^{2}U^{2}+sigma_{Y}^{3}U^{3}right)=mu_{Y}^{3}+3mu_{Y}sigma_{Y}^{2}$

  • $mathbb{E}Y^{4}=mathbb{E}left(mu_{Y}^{4}+4mu_{Y}^{3}sigma_{Y}U+6mu_{Y}^{2}sigma_{Y}^{2}U^{2}+4mu_{Y}sigma_{Y}^{3}U^{3}+sigma_{Y}^{4}U^{4}right)=mu_{Y}^{4}+6mu_{Y}^{2}sigma_{Y}^{2}+3sigma_{Y}^{4}$


Also have a look here for that.



Substituting we find the following expression for $mathsf{Var}XYleft(1-Yright)$:



$$left[sigma_{X}^{2}sigma_{Y}^{2}+sigma_{X}^{2}mu_{Y}^{2}+sigma_{Y}^{2}mu_{X}^{2}right]-2left[sigma_{X}^{2}mu_{Y}^{3}+3sigma_{X}^{2}sigma_{Y}^{2}mu_{Y}+2sigma_{Y}^{2}mu_{X}^{2}mu_{Y}right]+left[sigma_{X}^{2}mu_{Y}^{4}+6sigma_{X}^{2}sigma_{Y}^{2}mu_{Y}^{2}+3sigma_{X}^{2}sigma_{Y}^{4}+4sigma_{Y}^{2}mu_{X}^{2}mu_{Y}^{2}+2sigma_{Y}^{4}mu_{X}^{2}right]$$



I hope I did not make any mistakes (check me on it).






share|cite|improve this answer























  • This is wonderful, thank you! I wouldn't have thought of taking the covariance first. I'm not actually sure about the exact distributions of X and Y (they will be empirically determined). They'll probably end up Gaussian. Once I know them, then I would just have to do the integrals. Thanks so much, this is really helpful.
    – SabrinaChoice
    Dec 24 '18 at 14:47








  • 1




    @SabrinaChoice: In general, you can try to look at the moments of any sufficiently nice distribution if they exist, and each moment is a separate 'degree of freedom'. In fact, if all finite moments exist, then they uniquely characterize the distribution (if it has a continuous CDF). So we can also compare distributions by comparing their moments.
    – user21820
    Dec 24 '18 at 15:02






  • 1




    @SabrinaChoice I let you know that I found a mistake (and repaired). $mathbb EX^2mathbb EX^2$ was accidently replaced by $sigma_X^2+mu_X^2$, but had to be replaced by $(sigma_X^2+mu_X^2)(sigma_Y^2+mu_Y^2)$ of course.
    – drhab
    Dec 25 '18 at 16:44








  • 1




    In my former comment $mathbb EX^2mathbb EX^2$ should be $mathbb EX^2mathbb EY^2$. Also I edited for the special case that $Y$ takes normal distribution.
    – drhab
    Dec 26 '18 at 8:15












  • @drhab I finally had the chance to go through this carefully and check for errors. I couldn't find any. Thank you very much, I really appreciate it. My math skills are weak, but I feel like this particular example will make me much better equipped to solve a problem like this on my own next time I encounter one.
    – SabrinaChoice
    13 hours ago


















2














$begin{aligned}mathsf{Var}XYleft(1-Yright) & =mathsf{Covar}left(XYleft(1-Yright),XYleft(1-Yright)right)\
& =mathsf{Covar}left(XY-XY^{2},XY-XY^{2}right)\
& =mathsf{Var}left(XYright)-2mathsf{Covar}left(XY^{2},XYright)+mathsf{Var}left(XY^{2}right)\
& =left[mathbb{E}X^{2}Y^{2}-left(mathbb{E}XYright)^{2}right]-2left[mathbb{E}X^{2}Y^{3}-mathbb{E}XY^{2}mathbb{E}XYright]+left[mathbb{E}X^{2}Y^{4}-left(mathbb{E}XY^{2}right)^{2}right]\
& =left[mathbb{E}X^{2}mathbb{E}Y^{2}-left(mathbb{E}Xmathbb{E}Yright)^{2}right]-2left[mathbb{E}X^{2}mathbb{E}Y^{3}-mathbb{E}Xmathbb{E}Y^{2}mathbb{E}Xmathbb{E}Yright]+left[mathbb{E}X^{2}mathbb{E}Y^{4}-left(mathbb{E}Xmathbb{E}Y^{2}right)^{2}right]\
& =left[mathbb{E}X^{2}mathbb{E}Y^{2}-mu_{X}^{2}mu_{Y}^{2}right]-2left[mathbb{E}X^{2}mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}mathbb{E}Y^{2}right]+left[mathbb{E}X^{2}mathbb{E}Y^{4}-mu_{X}^{2}left(mathbb{E}Y^{2}right)^{2}right]\
& =left[left(sigma_{X}^{2}+mu_{X}^{2}right)left(sigma_{Y}^{2}+mu_{Y}^{2}right)-mu_{X}^{2}mu_{Y}^{2}right]-2left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}left(sigma_{Y}^{2}+mu_{Y}^{2}right)right]+left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{4}-mu_{X}^{2}left(sigma_{Y}^{2}+mu_{Y}^{2}right)^{2}right]\
& =left[sigma_{X}^{2}sigma_{Y}^{2}+sigma_{X}^{2}mu_{Y}^{2}+sigma_{Y}^{2}mu_{X}^{2}right]-2left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}left(sigma_{Y}^{2}+mu_{Y}^{2}right)right]+left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{4}-mu_{X}^{2}left(sigma_{Y}^{2}+mu_{Y}^{2}right)^{2}right]
end{aligned}
$



Especially the $5$-th equality rests on independence of $X$ and $Y$.



We cannot get any further since for the expressions $mathbb EY^3$ and $mathbb EY^4$ more information is needed concerning the distribution of $Y$.




edit:



Special case: $Ysimmathsf{Normal}left(mu_{Y},sigma_{Y}^{2}right)$.



Then $Y=mu_{Y}+sigma_{Y}U$ where $Usimmathsf{Normal}left(0,1right)$
and based on the knowledge that $mathbb{E}U^{3}=0$ and $mathbb{E}U^{4}=3$
we find:




  • $mathbb{E}Y^{3}=mathbb{E}left(mu_{Y}^{3}+3mu_{Y}^{2}sigma_{Y}U+3mu_{Y}sigma_{Y}^{2}U^{2}+sigma_{Y}^{3}U^{3}right)=mu_{Y}^{3}+3mu_{Y}sigma_{Y}^{2}$

  • $mathbb{E}Y^{4}=mathbb{E}left(mu_{Y}^{4}+4mu_{Y}^{3}sigma_{Y}U+6mu_{Y}^{2}sigma_{Y}^{2}U^{2}+4mu_{Y}sigma_{Y}^{3}U^{3}+sigma_{Y}^{4}U^{4}right)=mu_{Y}^{4}+6mu_{Y}^{2}sigma_{Y}^{2}+3sigma_{Y}^{4}$


Also have a look here for that.



Substituting we find the following expression for $mathsf{Var}XYleft(1-Yright)$:



$$left[sigma_{X}^{2}sigma_{Y}^{2}+sigma_{X}^{2}mu_{Y}^{2}+sigma_{Y}^{2}mu_{X}^{2}right]-2left[sigma_{X}^{2}mu_{Y}^{3}+3sigma_{X}^{2}sigma_{Y}^{2}mu_{Y}+2sigma_{Y}^{2}mu_{X}^{2}mu_{Y}right]+left[sigma_{X}^{2}mu_{Y}^{4}+6sigma_{X}^{2}sigma_{Y}^{2}mu_{Y}^{2}+3sigma_{X}^{2}sigma_{Y}^{4}+4sigma_{Y}^{2}mu_{X}^{2}mu_{Y}^{2}+2sigma_{Y}^{4}mu_{X}^{2}right]$$



I hope I did not make any mistakes (check me on it).






share|cite|improve this answer























  • This is wonderful, thank you! I wouldn't have thought of taking the covariance first. I'm not actually sure about the exact distributions of X and Y (they will be empirically determined). They'll probably end up Gaussian. Once I know them, then I would just have to do the integrals. Thanks so much, this is really helpful.
    – SabrinaChoice
    Dec 24 '18 at 14:47








  • 1




    @SabrinaChoice: In general, you can try to look at the moments of any sufficiently nice distribution if they exist, and each moment is a separate 'degree of freedom'. In fact, if all finite moments exist, then they uniquely characterize the distribution (if it has a continuous CDF). So we can also compare distributions by comparing their moments.
    – user21820
    Dec 24 '18 at 15:02






  • 1




    @SabrinaChoice I let you know that I found a mistake (and repaired). $mathbb EX^2mathbb EX^2$ was accidently replaced by $sigma_X^2+mu_X^2$, but had to be replaced by $(sigma_X^2+mu_X^2)(sigma_Y^2+mu_Y^2)$ of course.
    – drhab
    Dec 25 '18 at 16:44








  • 1




    In my former comment $mathbb EX^2mathbb EX^2$ should be $mathbb EX^2mathbb EY^2$. Also I edited for the special case that $Y$ takes normal distribution.
    – drhab
    Dec 26 '18 at 8:15












  • @drhab I finally had the chance to go through this carefully and check for errors. I couldn't find any. Thank you very much, I really appreciate it. My math skills are weak, but I feel like this particular example will make me much better equipped to solve a problem like this on my own next time I encounter one.
    – SabrinaChoice
    13 hours ago
















2












2








2






$begin{aligned}mathsf{Var}XYleft(1-Yright) & =mathsf{Covar}left(XYleft(1-Yright),XYleft(1-Yright)right)\
& =mathsf{Covar}left(XY-XY^{2},XY-XY^{2}right)\
& =mathsf{Var}left(XYright)-2mathsf{Covar}left(XY^{2},XYright)+mathsf{Var}left(XY^{2}right)\
& =left[mathbb{E}X^{2}Y^{2}-left(mathbb{E}XYright)^{2}right]-2left[mathbb{E}X^{2}Y^{3}-mathbb{E}XY^{2}mathbb{E}XYright]+left[mathbb{E}X^{2}Y^{4}-left(mathbb{E}XY^{2}right)^{2}right]\
& =left[mathbb{E}X^{2}mathbb{E}Y^{2}-left(mathbb{E}Xmathbb{E}Yright)^{2}right]-2left[mathbb{E}X^{2}mathbb{E}Y^{3}-mathbb{E}Xmathbb{E}Y^{2}mathbb{E}Xmathbb{E}Yright]+left[mathbb{E}X^{2}mathbb{E}Y^{4}-left(mathbb{E}Xmathbb{E}Y^{2}right)^{2}right]\
& =left[mathbb{E}X^{2}mathbb{E}Y^{2}-mu_{X}^{2}mu_{Y}^{2}right]-2left[mathbb{E}X^{2}mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}mathbb{E}Y^{2}right]+left[mathbb{E}X^{2}mathbb{E}Y^{4}-mu_{X}^{2}left(mathbb{E}Y^{2}right)^{2}right]\
& =left[left(sigma_{X}^{2}+mu_{X}^{2}right)left(sigma_{Y}^{2}+mu_{Y}^{2}right)-mu_{X}^{2}mu_{Y}^{2}right]-2left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}left(sigma_{Y}^{2}+mu_{Y}^{2}right)right]+left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{4}-mu_{X}^{2}left(sigma_{Y}^{2}+mu_{Y}^{2}right)^{2}right]\
& =left[sigma_{X}^{2}sigma_{Y}^{2}+sigma_{X}^{2}mu_{Y}^{2}+sigma_{Y}^{2}mu_{X}^{2}right]-2left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}left(sigma_{Y}^{2}+mu_{Y}^{2}right)right]+left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{4}-mu_{X}^{2}left(sigma_{Y}^{2}+mu_{Y}^{2}right)^{2}right]
end{aligned}
$



Especially the $5$-th equality rests on independence of $X$ and $Y$.



We cannot get any further since for the expressions $mathbb EY^3$ and $mathbb EY^4$ more information is needed concerning the distribution of $Y$.




edit:



Special case: $Ysimmathsf{Normal}left(mu_{Y},sigma_{Y}^{2}right)$.



Then $Y=mu_{Y}+sigma_{Y}U$ where $Usimmathsf{Normal}left(0,1right)$
and based on the knowledge that $mathbb{E}U^{3}=0$ and $mathbb{E}U^{4}=3$
we find:




  • $mathbb{E}Y^{3}=mathbb{E}left(mu_{Y}^{3}+3mu_{Y}^{2}sigma_{Y}U+3mu_{Y}sigma_{Y}^{2}U^{2}+sigma_{Y}^{3}U^{3}right)=mu_{Y}^{3}+3mu_{Y}sigma_{Y}^{2}$

  • $mathbb{E}Y^{4}=mathbb{E}left(mu_{Y}^{4}+4mu_{Y}^{3}sigma_{Y}U+6mu_{Y}^{2}sigma_{Y}^{2}U^{2}+4mu_{Y}sigma_{Y}^{3}U^{3}+sigma_{Y}^{4}U^{4}right)=mu_{Y}^{4}+6mu_{Y}^{2}sigma_{Y}^{2}+3sigma_{Y}^{4}$


Also have a look here for that.



Substituting we find the following expression for $mathsf{Var}XYleft(1-Yright)$:



$$left[sigma_{X}^{2}sigma_{Y}^{2}+sigma_{X}^{2}mu_{Y}^{2}+sigma_{Y}^{2}mu_{X}^{2}right]-2left[sigma_{X}^{2}mu_{Y}^{3}+3sigma_{X}^{2}sigma_{Y}^{2}mu_{Y}+2sigma_{Y}^{2}mu_{X}^{2}mu_{Y}right]+left[sigma_{X}^{2}mu_{Y}^{4}+6sigma_{X}^{2}sigma_{Y}^{2}mu_{Y}^{2}+3sigma_{X}^{2}sigma_{Y}^{4}+4sigma_{Y}^{2}mu_{X}^{2}mu_{Y}^{2}+2sigma_{Y}^{4}mu_{X}^{2}right]$$



I hope I did not make any mistakes (check me on it).






share|cite|improve this answer














$begin{aligned}mathsf{Var}XYleft(1-Yright) & =mathsf{Covar}left(XYleft(1-Yright),XYleft(1-Yright)right)\
& =mathsf{Covar}left(XY-XY^{2},XY-XY^{2}right)\
& =mathsf{Var}left(XYright)-2mathsf{Covar}left(XY^{2},XYright)+mathsf{Var}left(XY^{2}right)\
& =left[mathbb{E}X^{2}Y^{2}-left(mathbb{E}XYright)^{2}right]-2left[mathbb{E}X^{2}Y^{3}-mathbb{E}XY^{2}mathbb{E}XYright]+left[mathbb{E}X^{2}Y^{4}-left(mathbb{E}XY^{2}right)^{2}right]\
& =left[mathbb{E}X^{2}mathbb{E}Y^{2}-left(mathbb{E}Xmathbb{E}Yright)^{2}right]-2left[mathbb{E}X^{2}mathbb{E}Y^{3}-mathbb{E}Xmathbb{E}Y^{2}mathbb{E}Xmathbb{E}Yright]+left[mathbb{E}X^{2}mathbb{E}Y^{4}-left(mathbb{E}Xmathbb{E}Y^{2}right)^{2}right]\
& =left[mathbb{E}X^{2}mathbb{E}Y^{2}-mu_{X}^{2}mu_{Y}^{2}right]-2left[mathbb{E}X^{2}mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}mathbb{E}Y^{2}right]+left[mathbb{E}X^{2}mathbb{E}Y^{4}-mu_{X}^{2}left(mathbb{E}Y^{2}right)^{2}right]\
& =left[left(sigma_{X}^{2}+mu_{X}^{2}right)left(sigma_{Y}^{2}+mu_{Y}^{2}right)-mu_{X}^{2}mu_{Y}^{2}right]-2left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}left(sigma_{Y}^{2}+mu_{Y}^{2}right)right]+left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{4}-mu_{X}^{2}left(sigma_{Y}^{2}+mu_{Y}^{2}right)^{2}right]\
& =left[sigma_{X}^{2}sigma_{Y}^{2}+sigma_{X}^{2}mu_{Y}^{2}+sigma_{Y}^{2}mu_{X}^{2}right]-2left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{3}-mu_{X}^{2}mu_{Y}left(sigma_{Y}^{2}+mu_{Y}^{2}right)right]+left[left(sigma_{X}^{2}+mu_{X}^{2}right)mathbb{E}Y^{4}-mu_{X}^{2}left(sigma_{Y}^{2}+mu_{Y}^{2}right)^{2}right]
end{aligned}
$



Especially the $5$-th equality rests on independence of $X$ and $Y$.



We cannot get any further since for the expressions $mathbb EY^3$ and $mathbb EY^4$ more information is needed concerning the distribution of $Y$.




edit:



Special case: $Ysimmathsf{Normal}left(mu_{Y},sigma_{Y}^{2}right)$.



Then $Y=mu_{Y}+sigma_{Y}U$ where $Usimmathsf{Normal}left(0,1right)$
and based on the knowledge that $mathbb{E}U^{3}=0$ and $mathbb{E}U^{4}=3$
we find:




  • $mathbb{E}Y^{3}=mathbb{E}left(mu_{Y}^{3}+3mu_{Y}^{2}sigma_{Y}U+3mu_{Y}sigma_{Y}^{2}U^{2}+sigma_{Y}^{3}U^{3}right)=mu_{Y}^{3}+3mu_{Y}sigma_{Y}^{2}$

  • $mathbb{E}Y^{4}=mathbb{E}left(mu_{Y}^{4}+4mu_{Y}^{3}sigma_{Y}U+6mu_{Y}^{2}sigma_{Y}^{2}U^{2}+4mu_{Y}sigma_{Y}^{3}U^{3}+sigma_{Y}^{4}U^{4}right)=mu_{Y}^{4}+6mu_{Y}^{2}sigma_{Y}^{2}+3sigma_{Y}^{4}$


Also have a look here for that.



Substituting we find the following expression for $mathsf{Var}XYleft(1-Yright)$:



$$left[sigma_{X}^{2}sigma_{Y}^{2}+sigma_{X}^{2}mu_{Y}^{2}+sigma_{Y}^{2}mu_{X}^{2}right]-2left[sigma_{X}^{2}mu_{Y}^{3}+3sigma_{X}^{2}sigma_{Y}^{2}mu_{Y}+2sigma_{Y}^{2}mu_{X}^{2}mu_{Y}right]+left[sigma_{X}^{2}mu_{Y}^{4}+6sigma_{X}^{2}sigma_{Y}^{2}mu_{Y}^{2}+3sigma_{X}^{2}sigma_{Y}^{4}+4sigma_{Y}^{2}mu_{X}^{2}mu_{Y}^{2}+2sigma_{Y}^{4}mu_{X}^{2}right]$$



I hope I did not make any mistakes (check me on it).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 12 hours ago

























answered Dec 24 '18 at 13:11









drhab

98.1k544129




98.1k544129












  • This is wonderful, thank you! I wouldn't have thought of taking the covariance first. I'm not actually sure about the exact distributions of X and Y (they will be empirically determined). They'll probably end up Gaussian. Once I know them, then I would just have to do the integrals. Thanks so much, this is really helpful.
    – SabrinaChoice
    Dec 24 '18 at 14:47








  • 1




    @SabrinaChoice: In general, you can try to look at the moments of any sufficiently nice distribution if they exist, and each moment is a separate 'degree of freedom'. In fact, if all finite moments exist, then they uniquely characterize the distribution (if it has a continuous CDF). So we can also compare distributions by comparing their moments.
    – user21820
    Dec 24 '18 at 15:02






  • 1




    @SabrinaChoice I let you know that I found a mistake (and repaired). $mathbb EX^2mathbb EX^2$ was accidently replaced by $sigma_X^2+mu_X^2$, but had to be replaced by $(sigma_X^2+mu_X^2)(sigma_Y^2+mu_Y^2)$ of course.
    – drhab
    Dec 25 '18 at 16:44








  • 1




    In my former comment $mathbb EX^2mathbb EX^2$ should be $mathbb EX^2mathbb EY^2$. Also I edited for the special case that $Y$ takes normal distribution.
    – drhab
    Dec 26 '18 at 8:15












  • @drhab I finally had the chance to go through this carefully and check for errors. I couldn't find any. Thank you very much, I really appreciate it. My math skills are weak, but I feel like this particular example will make me much better equipped to solve a problem like this on my own next time I encounter one.
    – SabrinaChoice
    13 hours ago




















  • This is wonderful, thank you! I wouldn't have thought of taking the covariance first. I'm not actually sure about the exact distributions of X and Y (they will be empirically determined). They'll probably end up Gaussian. Once I know them, then I would just have to do the integrals. Thanks so much, this is really helpful.
    – SabrinaChoice
    Dec 24 '18 at 14:47








  • 1




    @SabrinaChoice: In general, you can try to look at the moments of any sufficiently nice distribution if they exist, and each moment is a separate 'degree of freedom'. In fact, if all finite moments exist, then they uniquely characterize the distribution (if it has a continuous CDF). So we can also compare distributions by comparing their moments.
    – user21820
    Dec 24 '18 at 15:02






  • 1




    @SabrinaChoice I let you know that I found a mistake (and repaired). $mathbb EX^2mathbb EX^2$ was accidently replaced by $sigma_X^2+mu_X^2$, but had to be replaced by $(sigma_X^2+mu_X^2)(sigma_Y^2+mu_Y^2)$ of course.
    – drhab
    Dec 25 '18 at 16:44








  • 1




    In my former comment $mathbb EX^2mathbb EX^2$ should be $mathbb EX^2mathbb EY^2$. Also I edited for the special case that $Y$ takes normal distribution.
    – drhab
    Dec 26 '18 at 8:15












  • @drhab I finally had the chance to go through this carefully and check for errors. I couldn't find any. Thank you very much, I really appreciate it. My math skills are weak, but I feel like this particular example will make me much better equipped to solve a problem like this on my own next time I encounter one.
    – SabrinaChoice
    13 hours ago


















This is wonderful, thank you! I wouldn't have thought of taking the covariance first. I'm not actually sure about the exact distributions of X and Y (they will be empirically determined). They'll probably end up Gaussian. Once I know them, then I would just have to do the integrals. Thanks so much, this is really helpful.
– SabrinaChoice
Dec 24 '18 at 14:47






This is wonderful, thank you! I wouldn't have thought of taking the covariance first. I'm not actually sure about the exact distributions of X and Y (they will be empirically determined). They'll probably end up Gaussian. Once I know them, then I would just have to do the integrals. Thanks so much, this is really helpful.
– SabrinaChoice
Dec 24 '18 at 14:47






1




1




@SabrinaChoice: In general, you can try to look at the moments of any sufficiently nice distribution if they exist, and each moment is a separate 'degree of freedom'. In fact, if all finite moments exist, then they uniquely characterize the distribution (if it has a continuous CDF). So we can also compare distributions by comparing their moments.
– user21820
Dec 24 '18 at 15:02




@SabrinaChoice: In general, you can try to look at the moments of any sufficiently nice distribution if they exist, and each moment is a separate 'degree of freedom'. In fact, if all finite moments exist, then they uniquely characterize the distribution (if it has a continuous CDF). So we can also compare distributions by comparing their moments.
– user21820
Dec 24 '18 at 15:02




1




1




@SabrinaChoice I let you know that I found a mistake (and repaired). $mathbb EX^2mathbb EX^2$ was accidently replaced by $sigma_X^2+mu_X^2$, but had to be replaced by $(sigma_X^2+mu_X^2)(sigma_Y^2+mu_Y^2)$ of course.
– drhab
Dec 25 '18 at 16:44






@SabrinaChoice I let you know that I found a mistake (and repaired). $mathbb EX^2mathbb EX^2$ was accidently replaced by $sigma_X^2+mu_X^2$, but had to be replaced by $(sigma_X^2+mu_X^2)(sigma_Y^2+mu_Y^2)$ of course.
– drhab
Dec 25 '18 at 16:44






1




1




In my former comment $mathbb EX^2mathbb EX^2$ should be $mathbb EX^2mathbb EY^2$. Also I edited for the special case that $Y$ takes normal distribution.
– drhab
Dec 26 '18 at 8:15






In my former comment $mathbb EX^2mathbb EX^2$ should be $mathbb EX^2mathbb EY^2$. Also I edited for the special case that $Y$ takes normal distribution.
– drhab
Dec 26 '18 at 8:15














@drhab I finally had the chance to go through this carefully and check for errors. I couldn't find any. Thank you very much, I really appreciate it. My math skills are weak, but I feel like this particular example will make me much better equipped to solve a problem like this on my own next time I encounter one.
– SabrinaChoice
13 hours ago






@drhab I finally had the chance to go through this carefully and check for errors. I couldn't find any. Thank you very much, I really appreciate it. My math skills are weak, but I feel like this particular example will make me much better equipped to solve a problem like this on my own next time I encounter one.
– SabrinaChoice
13 hours ago




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050093%2fvariance-of-xy1-y-in-terms-of-the-means-and-variances-x-and-y%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Mario Kart Wii

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
Read more

The Binding of Isaac: Rebirth/Afterbirth

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
Read more

What does “Dominus providebit” mean?

Image
5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
Read more