For functions of an RV, why doesn’t equating PDFs work?
$begingroup$
Let $ X sim N(mu, sigma^2) $ and $Z = frac{X-mu}{sigma}$. Suppose I want to prove $f_Z(x) = frac{1}{sqrt{2pi}} exp(-frac{1}{2}x^2)$. I tried reasoning:
$$
f_Z(x) = P(Z=x) = P(frac{X-mu}{sigma} = x) = P(X = sigma x + mu) = f_X(sigma x + mu) \
= frac1{sqrt{2pi}} frac1{sigma} expleft(-frac{1}{2} left(frac{ (sigma x + mu) - mu }{sigma}right)^2 right) = frac1{sqrt{2pi}} frac1{sigma} expleft(-frac{1}{2} x^2 right)
$$
However, note the pesky $frac1sigma$ that still remains in the above expression. That shouldn’t be there. Why is this happening? Is it invalid to equate the PDFs as I did?
Edit: From the comments, I realized my blooper: $f_Z(x)$ is not in fact the same as $P(Z=x)$, because of course the latter is always zero for continuous variables like $X$ and $Z$!
My revised question would be: Is there any way to understand the distribution of a function of an RV without going through the CDF, some method that works using PDFs?
statistics random-variables
asked Jan 14 at 5:24
AlexAlex
11
$endgroup$
add a comment |
$begingroup$
Let $ X sim N(mu, sigma^2) $ and $Z = frac{X-mu}{sigma}$. Suppose I want to prove $f_Z(x) = frac{1}{sqrt{2pi}} exp(-frac{1}{2}x^2)$. I tried reasoning:
$$
f_Z(x) = P(Z=x) = P(frac{X-mu}{sigma} = x) = P(X = sigma x + mu) = f_X(sigma x + mu) \
= frac1{sqrt{2pi}} frac1{sigma} expleft(-frac{1}{2} left(frac{ (sigma x + mu) - mu }{sigma}right)^2 right) = frac1{sqrt{2pi}} frac1{sigma} expleft(-frac{1}{2} x^2 right)
$$
However, note the pesky $frac1sigma$ that still remains in the above expression. That shouldn’t be there. Why is this happening? Is it invalid to equate the PDFs as I did?
Edit: From the comments, I realized my blooper: $f_Z(x)$ is not in fact the same as $P(Z=x)$, because of course the latter is always zero for continuous variables like $X$ and $Z$!
My revised question would be: Is there any way to understand the distribution of a function of an RV without going through the CDF, some method that works using PDFs?
statistics random-variables
asked Jan 14 at 5:24
AlexAlex
11
$endgroup$
1
$begingroup$
You are using arguments valid only for discrete RV's to continuous RV's. Here $P(X=x)=0$ for all $x$ and $P(Z=z)=0$ for all $z$.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 5:27
1
$begingroup$
$P(Z=x)=0$, not $F_Z(x)$.
$endgroup$
– Lord Shark the Unknown
Jan 14 at 5:27
$begingroup$
@KaviRamaMurthy Yes, what a brain blooper! I hope putting up this question might help someone equally dumbstruck in the future.
$endgroup$
– Alex
Jan 14 at 5:57
$begingroup$
You want to use a combination of (a) if $Y=X+a$ then $f_Y(y)=f_X(y-a)$ and (b) if $W=bX$ then $f_W(w)=frac1{|b|}f_X(w/b)$. While the first of these is in a sense obvious, the second needs an argument something like the CDF, or the integral of the density being $1$, or the use of the Jacobian
$endgroup$
– Henry
Jan 14 at 8:09
add a comment |
$begingroup$
Let $ X sim N(mu, sigma^2) $ and $Z = frac{X-mu}{sigma}$. Suppose I want to prove $f_Z(x) = frac{1}{sqrt{2pi}} exp(-frac{1}{2}x^2)$. I tried reasoning:
$$
f_Z(x) = P(Z=x) = P(frac{X-mu}{sigma} = x) = P(X = sigma x + mu) = f_X(sigma x + mu) \
= frac1{sqrt{2pi}} frac1{sigma} expleft(-frac{1}{2} left(frac{ (sigma x + mu) - mu }{sigma}right)^2 right) = frac1{sqrt{2pi}} frac1{sigma} expleft(-frac{1}{2} x^2 right)
$$
However, note the pesky $frac1sigma$ that still remains in the above expression. That shouldn’t be there. Why is this happening? Is it invalid to equate the PDFs as I did?
Edit: From the comments, I realized my blooper: $f_Z(x)$ is not in fact the same as $P(Z=x)$, because of course the latter is always zero for continuous variables like $X$ and $Z$!
My revised question would be: Is there any way to understand the distribution of a function of an RV without going through the CDF, some method that works using PDFs?
statistics random-variables
asked Jan 14 at 5:24
AlexAlex
11
$endgroup$
Let $ X sim N(mu, sigma^2) $ and $Z = frac{X-mu}{sigma}$. Suppose I want to prove $f_Z(x) = frac{1}{sqrt{2pi}} exp(-frac{1}{2}x^2)$. I tried reasoning:
$$
f_Z(x) = P(Z=x) = P(frac{X-mu}{sigma} = x) = P(X = sigma x + mu) = f_X(sigma x + mu) \
= frac1{sqrt{2pi}} frac1{sigma} expleft(-frac{1}{2} left(frac{ (sigma x + mu) - mu }{sigma}right)^2 right) = frac1{sqrt{2pi}} frac1{sigma} expleft(-frac{1}{2} x^2 right)
$$
However, note the pesky $frac1sigma$ that still remains in the above expression. That shouldn’t be there. Why is this happening? Is it invalid to equate the PDFs as I did?
Edit: From the comments, I realized my blooper: $f_Z(x)$ is not in fact the same as $P(Z=x)$, because of course the latter is always zero for continuous variables like $X$ and $Z$!
My revised question would be: Is there any way to understand the distribution of a function of an RV without going through the CDF, some method that works using PDFs?
statistics random-variables
statistics random-variables
asked Jan 14 at 5:24
AlexAlex
11
asked Jan 14 at 5:24
AlexAlex
11
edited Jan 14 at 6:00
Alex
asked Jan 14 at 5:24
AlexAlex
11
asked Jan 14 at 5:24
AlexAlex
11
asked Jan 14 at 5:24
AlexAlex
11
11
1
$begingroup$
You are using arguments valid only for discrete RV's to continuous RV's. Here $P(X=x)=0$ for all $x$ and $P(Z=z)=0$ for all $z$.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 5:27
1
$begingroup$
$P(Z=x)=0$, not $F_Z(x)$.
$endgroup$
– Lord Shark the Unknown
Jan 14 at 5:27
$begingroup$
@KaviRamaMurthy Yes, what a brain blooper! I hope putting up this question might help someone equally dumbstruck in the future.
$endgroup$
– Alex
Jan 14 at 5:57
$begingroup$
You want to use a combination of (a) if $Y=X+a$ then $f_Y(y)=f_X(y-a)$ and (b) if $W=bX$ then $f_W(w)=frac1{|b|}f_X(w/b)$. While the first of these is in a sense obvious, the second needs an argument something like the CDF, or the integral of the density being $1$, or the use of the Jacobian
$endgroup$
– Henry
Jan 14 at 8:09
add a comment |
1
$begingroup$
You are using arguments valid only for discrete RV's to continuous RV's. Here $P(X=x)=0$ for all $x$ and $P(Z=z)=0$ for all $z$.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 5:27
1
$begingroup$
$P(Z=x)=0$, not $F_Z(x)$.
$endgroup$
– Lord Shark the Unknown
Jan 14 at 5:27
$begingroup$
@KaviRamaMurthy Yes, what a brain blooper! I hope putting up this question might help someone equally dumbstruck in the future.
$endgroup$
– Alex
Jan 14 at 5:57
$begingroup$
You want to use a combination of (a) if $Y=X+a$ then $f_Y(y)=f_X(y-a)$ and (b) if $W=bX$ then $f_W(w)=frac1{|b|}f_X(w/b)$. While the first of these is in a sense obvious, the second needs an argument something like the CDF, or the integral of the density being $1$, or the use of the Jacobian
$endgroup$
– Henry
Jan 14 at 8:09
1
1
$begingroup$
You are using arguments valid only for discrete RV's to continuous RV's. Here $P(X=x)=0$ for all $x$ and $P(Z=z)=0$ for all $z$.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 5:27
$begingroup$
You are using arguments valid only for discrete RV's to continuous RV's. Here $P(X=x)=0$ for all $x$ and $P(Z=z)=0$ for all $z$.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 5:27
1
1
$begingroup$
$P(Z=x)=0$, not $F_Z(x)$.
$endgroup$
– Lord Shark the Unknown
Jan 14 at 5:27
$begingroup$
$P(Z=x)=0$, not $F_Z(x)$.
$endgroup$
– Lord Shark the Unknown
Jan 14 at 5:27
$begingroup$
@KaviRamaMurthy Yes, what a brain blooper! I hope putting up this question might help someone equally dumbstruck in the future.
$endgroup$
– Alex
Jan 14 at 5:57
$begingroup$
@KaviRamaMurthy Yes, what a brain blooper! I hope putting up this question might help someone equally dumbstruck in the future.
$endgroup$
– Alex
Jan 14 at 5:57
$begingroup$
You want to use a combination of (a) if $Y=X+a$ then $f_Y(y)=f_X(y-a)$ and (b) if $W=bX$ then $f_W(w)=frac1{|b|}f_X(w/b)$. While the first of these is in a sense obvious, the second needs an argument something like the CDF, or the integral of the density being $1$, or the use of the Jacobian
$endgroup$
– Henry
Jan 14 at 8:09
$begingroup$
You want to use a combination of (a) if $Y=X+a$ then $f_Y(y)=f_X(y-a)$ and (b) if $W=bX$ then $f_W(w)=frac1{|b|}f_X(w/b)$. While the first of these is in a sense obvious, the second needs an argument something like the CDF, or the integral of the density being $1$, or the use of the Jacobian
$endgroup$
– Henry
Jan 14 at 8:09
add a comment |
0
active
oldest
votes
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3072880%2ffor-functions-of-an-rv-why-doesn-t-equating-pdfs-work%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3072880%2ffor-functions-of-an-rv-why-doesn-t-equating-pdfs-work%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
1
$begingroup$
You are using arguments valid only for discrete RV's to continuous RV's. Here $P(X=x)=0$ for all $x$ and $P(Z=z)=0$ for all $z$.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 5:27
1
$begingroup$
$P(Z=x)=0$, not $F_Z(x)$.
$endgroup$
– Lord Shark the Unknown
Jan 14 at 5:27
$begingroup$
@KaviRamaMurthy Yes, what a brain blooper! I hope putting up this question might help someone equally dumbstruck in the future.
$endgroup$
– Alex
Jan 14 at 5:57
$begingroup$
You want to use a combination of (a) if $Y=X+a$ then $f_Y(y)=f_X(y-a)$ and (b) if $W=bX$ then $f_W(w)=frac1{|b|}f_X(w/b)$. While the first of these is in a sense obvious, the second needs an argument something like the CDF, or the integral of the density being $1$, or the use of the Jacobian
$endgroup$
– Henry
Jan 14 at 8:09