A experiment of throwing two fair cubic dice
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Two fair cubic dice are thrown repeatedly in an experiment. Let $X_i$ be the absolute difference of the value of the two dice in the $i$-th throw. The experiment will be stopped when $X_i=0$. Let $Y=sum X_i$. How can I evaluate the mean and variance of Y, if the experiment is stopped at the $n$-th throw? And if the experiment is stopped at the 4-th throw and $Y=5$, what is the probability of $X_1=1$?
probability
asked Jan 22 at 8:09
Weihao HuangWeihao Huang
112
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add a comment |
$begingroup$
Two fair cubic dice are thrown repeatedly in an experiment. Let $X_i$ be the absolute difference of the value of the two dice in the $i$-th throw. The experiment will be stopped when $X_i=0$. Let $Y=sum X_i$. How can I evaluate the mean and variance of Y, if the experiment is stopped at the $n$-th throw? And if the experiment is stopped at the 4-th throw and $Y=5$, what is the probability of $X_1=1$?
probability
asked Jan 22 at 8:09
Weihao HuangWeihao Huang
112
$endgroup$
add a comment |
$begingroup$
Two fair cubic dice are thrown repeatedly in an experiment. Let $X_i$ be the absolute difference of the value of the two dice in the $i$-th throw. The experiment will be stopped when $X_i=0$. Let $Y=sum X_i$. How can I evaluate the mean and variance of Y, if the experiment is stopped at the $n$-th throw? And if the experiment is stopped at the 4-th throw and $Y=5$, what is the probability of $X_1=1$?
probability
asked Jan 22 at 8:09
Weihao HuangWeihao Huang
112
$endgroup$
Two fair cubic dice are thrown repeatedly in an experiment. Let $X_i$ be the absolute difference of the value of the two dice in the $i$-th throw. The experiment will be stopped when $X_i=0$. Let $Y=sum X_i$. How can I evaluate the mean and variance of Y, if the experiment is stopped at the $n$-th throw? And if the experiment is stopped at the 4-th throw and $Y=5$, what is the probability of $X_1=1$?
probability
probability
asked Jan 22 at 8:09
Weihao HuangWeihao Huang
112
asked Jan 22 at 8:09
Weihao HuangWeihao Huang
112
asked Jan 22 at 8:09
Weihao HuangWeihao Huang
112
asked Jan 22 at 8:09
Weihao HuangWeihao Huang
112
asked Jan 22 at 8:09
Weihao HuangWeihao Huang
112
112
add a comment |
add a comment |
1 Answer
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oldest
votes
$begingroup$
Let $E_n$ denote the event that the experiment is stopped at the $n$-th throw.
Then to be found is:$$mathbb E[Ymid E_n]=sum_{i=1}^{n-1}mathbb E[X_imid E_n]=(n-1)mathbb E[X_1mid E_n]=(n-1)mathbb E[X_1mid X_1>0]$$
For the last equality observe that $E_n={X_1>0}cap F_n$ where $F_n:={X_2>0,cdots,X_{n-1}>0,X_n=0}$ is an event such that $mathbf1_{F_n}$ and $X_1$ are independent so that:$$mathbb E[X_1mid E_n]=frac{mathbb EX_1mathbf1_{X_1>0}mathbf1_{F_n}}{mathbb Emathbf1_{X_1>0}mathbf1_{F_n}}=frac{mathbb EX_1mathbf1_{X_1>0}mathbb Emathbf1_{F_n}}{mathbb Emathbf1_{X_1>0}mathbb Emathbf1_{F_n}}=frac{mathbb EX_1mathbf1_{X_1>0}}{mathbb Emathbf1_{X_1>0}}=mathbb E[X_1mid X_1>0]$$
To find the mean of $Y$ it is enough now to find $mathbb E[X_1mid X_1>0]$ and I leave that to you.
Similarly we can find: $$mathbb E[Y^2mid E_n]=$$$$sum_{i=1}^{n-1}sum_{j=1}^{n-1}mathbb E[X_iX_jmid E_n]=(n-1)(n-2)mathbb E[X_1X_2mid X_1>0,X_2>0]+(n-1)mathbb E[X_1^2mid X_1>0]$$and then variance $mathbb E[Y^2mid E_n]-(mathbb E[Ymid E_n])^2$.
To be found is: $$P(X_1=1mid X_1>0,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)}{P(X_1>0,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)}=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_2+X_3=4)P(X_4=0)}{P(X_1>0,X_2>0,X_3>0,X_1+X_2+X_3=5)P(X_4=0)}=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_2+X_3=4)}{P(X_1>0,X_2>0,X_3>0,X_1+X_2+X_3=5)}=$$$$frac{sum_{s,t>0,s+t=4}P(X_1=1)P(X_2=s)P(X_3=t)}{sum_{r,s,t>0,r+s+t=5}P(X_1=r)P(X_2=s)P(X_3=t)}$$
answered Jan 22 at 9:58
drhabdrhab
102k545136
$endgroup$
$begingroup$
Thanks for the reply. Since I not good at probability, I hope you can confirm my works. The mean is $E(Y)=(n-1)E(X_i)$ and the variance is $Var(Y)=(n-1)[E(X_i^2)-(E(X_i))^2]$. I don't understand why you write $X_1>0$ as a condition since the value of $X_i$ has already grater than 0.
$endgroup$
– Weihao Huang
Jan 22 at 10:48
$begingroup$
Do you mean that $X_i>0$ (where $i>1$) already implies that $X_1>0$? I do that because I look at this as a process that does not stop (that does not harm, $Y$ can still be defined properly), and work with $X_i$ that are independent. If e.g. conditions like $X_2>0implies X_1>0$ are build in (I avoid that) then this (valuable) independence is lost.
$endgroup$
– drhab
Jan 22 at 11:26
$begingroup$
I see, it not that complicated. $X_i$ is defined as the $|D_1-D_2|$, so it must be greater than $0$.
$endgroup$
– Weihao Huang
Jan 22 at 11:33
$begingroup$
"..so it must be greater than $0$." What do you mean? For every index $i$ it is possible that $D_1^{(i)}=D_2^{(i)}$ or equivalently $X_i=0$.
$endgroup$
– drhab
Jan 22 at 11:38
$begingroup$
I see what you mean, thanks
$endgroup$
– Weihao Huang
Jan 22 at 11:59
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $E_n$ denote the event that the experiment is stopped at the $n$-th throw.
Then to be found is:$$mathbb E[Ymid E_n]=sum_{i=1}^{n-1}mathbb E[X_imid E_n]=(n-1)mathbb E[X_1mid E_n]=(n-1)mathbb E[X_1mid X_1>0]$$
For the last equality observe that $E_n={X_1>0}cap F_n$ where $F_n:={X_2>0,cdots,X_{n-1}>0,X_n=0}$ is an event such that $mathbf1_{F_n}$ and $X_1$ are independent so that:$$mathbb E[X_1mid E_n]=frac{mathbb EX_1mathbf1_{X_1>0}mathbf1_{F_n}}{mathbb Emathbf1_{X_1>0}mathbf1_{F_n}}=frac{mathbb EX_1mathbf1_{X_1>0}mathbb Emathbf1_{F_n}}{mathbb Emathbf1_{X_1>0}mathbb Emathbf1_{F_n}}=frac{mathbb EX_1mathbf1_{X_1>0}}{mathbb Emathbf1_{X_1>0}}=mathbb E[X_1mid X_1>0]$$
To find the mean of $Y$ it is enough now to find $mathbb E[X_1mid X_1>0]$ and I leave that to you.
Similarly we can find: $$mathbb E[Y^2mid E_n]=$$$$sum_{i=1}^{n-1}sum_{j=1}^{n-1}mathbb E[X_iX_jmid E_n]=(n-1)(n-2)mathbb E[X_1X_2mid X_1>0,X_2>0]+(n-1)mathbb E[X_1^2mid X_1>0]$$and then variance $mathbb E[Y^2mid E_n]-(mathbb E[Ymid E_n])^2$.
To be found is: $$P(X_1=1mid X_1>0,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)}{P(X_1>0,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)}=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_2+X_3=4)P(X_4=0)}{P(X_1>0,X_2>0,X_3>0,X_1+X_2+X_3=5)P(X_4=0)}=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_2+X_3=4)}{P(X_1>0,X_2>0,X_3>0,X_1+X_2+X_3=5)}=$$$$frac{sum_{s,t>0,s+t=4}P(X_1=1)P(X_2=s)P(X_3=t)}{sum_{r,s,t>0,r+s+t=5}P(X_1=r)P(X_2=s)P(X_3=t)}$$
answered Jan 22 at 9:58
drhabdrhab
102k545136
$endgroup$
$begingroup$
Thanks for the reply. Since I not good at probability, I hope you can confirm my works. The mean is $E(Y)=(n-1)E(X_i)$ and the variance is $Var(Y)=(n-1)[E(X_i^2)-(E(X_i))^2]$. I don't understand why you write $X_1>0$ as a condition since the value of $X_i$ has already grater than 0.
$endgroup$
– Weihao Huang
Jan 22 at 10:48
$begingroup$
Do you mean that $X_i>0$ (where $i>1$) already implies that $X_1>0$? I do that because I look at this as a process that does not stop (that does not harm, $Y$ can still be defined properly), and work with $X_i$ that are independent. If e.g. conditions like $X_2>0implies X_1>0$ are build in (I avoid that) then this (valuable) independence is lost.
$endgroup$
– drhab
Jan 22 at 11:26
$begingroup$
I see, it not that complicated. $X_i$ is defined as the $|D_1-D_2|$, so it must be greater than $0$.
$endgroup$
– Weihao Huang
Jan 22 at 11:33
$begingroup$
"..so it must be greater than $0$." What do you mean? For every index $i$ it is possible that $D_1^{(i)}=D_2^{(i)}$ or equivalently $X_i=0$.
$endgroup$
– drhab
Jan 22 at 11:38
$begingroup$
I see what you mean, thanks
$endgroup$
– Weihao Huang
Jan 22 at 11:59
add a comment |
$begingroup$
Let $E_n$ denote the event that the experiment is stopped at the $n$-th throw.
Then to be found is:$$mathbb E[Ymid E_n]=sum_{i=1}^{n-1}mathbb E[X_imid E_n]=(n-1)mathbb E[X_1mid E_n]=(n-1)mathbb E[X_1mid X_1>0]$$
For the last equality observe that $E_n={X_1>0}cap F_n$ where $F_n:={X_2>0,cdots,X_{n-1}>0,X_n=0}$ is an event such that $mathbf1_{F_n}$ and $X_1$ are independent so that:$$mathbb E[X_1mid E_n]=frac{mathbb EX_1mathbf1_{X_1>0}mathbf1_{F_n}}{mathbb Emathbf1_{X_1>0}mathbf1_{F_n}}=frac{mathbb EX_1mathbf1_{X_1>0}mathbb Emathbf1_{F_n}}{mathbb Emathbf1_{X_1>0}mathbb Emathbf1_{F_n}}=frac{mathbb EX_1mathbf1_{X_1>0}}{mathbb Emathbf1_{X_1>0}}=mathbb E[X_1mid X_1>0]$$
To find the mean of $Y$ it is enough now to find $mathbb E[X_1mid X_1>0]$ and I leave that to you.
Similarly we can find: $$mathbb E[Y^2mid E_n]=$$$$sum_{i=1}^{n-1}sum_{j=1}^{n-1}mathbb E[X_iX_jmid E_n]=(n-1)(n-2)mathbb E[X_1X_2mid X_1>0,X_2>0]+(n-1)mathbb E[X_1^2mid X_1>0]$$and then variance $mathbb E[Y^2mid E_n]-(mathbb E[Ymid E_n])^2$.
To be found is: $$P(X_1=1mid X_1>0,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)}{P(X_1>0,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)}=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_2+X_3=4)P(X_4=0)}{P(X_1>0,X_2>0,X_3>0,X_1+X_2+X_3=5)P(X_4=0)}=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_2+X_3=4)}{P(X_1>0,X_2>0,X_3>0,X_1+X_2+X_3=5)}=$$$$frac{sum_{s,t>0,s+t=4}P(X_1=1)P(X_2=s)P(X_3=t)}{sum_{r,s,t>0,r+s+t=5}P(X_1=r)P(X_2=s)P(X_3=t)}$$
answered Jan 22 at 9:58
drhabdrhab
102k545136
$endgroup$
$begingroup$
Thanks for the reply. Since I not good at probability, I hope you can confirm my works. The mean is $E(Y)=(n-1)E(X_i)$ and the variance is $Var(Y)=(n-1)[E(X_i^2)-(E(X_i))^2]$. I don't understand why you write $X_1>0$ as a condition since the value of $X_i$ has already grater than 0.
$endgroup$
– Weihao Huang
Jan 22 at 10:48
$begingroup$
Do you mean that $X_i>0$ (where $i>1$) already implies that $X_1>0$? I do that because I look at this as a process that does not stop (that does not harm, $Y$ can still be defined properly), and work with $X_i$ that are independent. If e.g. conditions like $X_2>0implies X_1>0$ are build in (I avoid that) then this (valuable) independence is lost.
$endgroup$
– drhab
Jan 22 at 11:26
$begingroup$
I see, it not that complicated. $X_i$ is defined as the $|D_1-D_2|$, so it must be greater than $0$.
$endgroup$
– Weihao Huang
Jan 22 at 11:33
$begingroup$
"..so it must be greater than $0$." What do you mean? For every index $i$ it is possible that $D_1^{(i)}=D_2^{(i)}$ or equivalently $X_i=0$.
$endgroup$
– drhab
Jan 22 at 11:38
$begingroup$
I see what you mean, thanks
$endgroup$
– Weihao Huang
Jan 22 at 11:59
add a comment |
$begingroup$
Let $E_n$ denote the event that the experiment is stopped at the $n$-th throw.
Then to be found is:$$mathbb E[Ymid E_n]=sum_{i=1}^{n-1}mathbb E[X_imid E_n]=(n-1)mathbb E[X_1mid E_n]=(n-1)mathbb E[X_1mid X_1>0]$$
For the last equality observe that $E_n={X_1>0}cap F_n$ where $F_n:={X_2>0,cdots,X_{n-1}>0,X_n=0}$ is an event such that $mathbf1_{F_n}$ and $X_1$ are independent so that:$$mathbb E[X_1mid E_n]=frac{mathbb EX_1mathbf1_{X_1>0}mathbf1_{F_n}}{mathbb Emathbf1_{X_1>0}mathbf1_{F_n}}=frac{mathbb EX_1mathbf1_{X_1>0}mathbb Emathbf1_{F_n}}{mathbb Emathbf1_{X_1>0}mathbb Emathbf1_{F_n}}=frac{mathbb EX_1mathbf1_{X_1>0}}{mathbb Emathbf1_{X_1>0}}=mathbb E[X_1mid X_1>0]$$
To find the mean of $Y$ it is enough now to find $mathbb E[X_1mid X_1>0]$ and I leave that to you.
Similarly we can find: $$mathbb E[Y^2mid E_n]=$$$$sum_{i=1}^{n-1}sum_{j=1}^{n-1}mathbb E[X_iX_jmid E_n]=(n-1)(n-2)mathbb E[X_1X_2mid X_1>0,X_2>0]+(n-1)mathbb E[X_1^2mid X_1>0]$$and then variance $mathbb E[Y^2mid E_n]-(mathbb E[Ymid E_n])^2$.
To be found is: $$P(X_1=1mid X_1>0,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)}{P(X_1>0,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)}=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_2+X_3=4)P(X_4=0)}{P(X_1>0,X_2>0,X_3>0,X_1+X_2+X_3=5)P(X_4=0)}=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_2+X_3=4)}{P(X_1>0,X_2>0,X_3>0,X_1+X_2+X_3=5)}=$$$$frac{sum_{s,t>0,s+t=4}P(X_1=1)P(X_2=s)P(X_3=t)}{sum_{r,s,t>0,r+s+t=5}P(X_1=r)P(X_2=s)P(X_3=t)}$$
answered Jan 22 at 9:58
drhabdrhab
102k545136
$endgroup$
Let $E_n$ denote the event that the experiment is stopped at the $n$-th throw.
Then to be found is:$$mathbb E[Ymid E_n]=sum_{i=1}^{n-1}mathbb E[X_imid E_n]=(n-1)mathbb E[X_1mid E_n]=(n-1)mathbb E[X_1mid X_1>0]$$
For the last equality observe that $E_n={X_1>0}cap F_n$ where $F_n:={X_2>0,cdots,X_{n-1}>0,X_n=0}$ is an event such that $mathbf1_{F_n}$ and $X_1$ are independent so that:$$mathbb E[X_1mid E_n]=frac{mathbb EX_1mathbf1_{X_1>0}mathbf1_{F_n}}{mathbb Emathbf1_{X_1>0}mathbf1_{F_n}}=frac{mathbb EX_1mathbf1_{X_1>0}mathbb Emathbf1_{F_n}}{mathbb Emathbf1_{X_1>0}mathbb Emathbf1_{F_n}}=frac{mathbb EX_1mathbf1_{X_1>0}}{mathbb Emathbf1_{X_1>0}}=mathbb E[X_1mid X_1>0]$$
To find the mean of $Y$ it is enough now to find $mathbb E[X_1mid X_1>0]$ and I leave that to you.
Similarly we can find: $$mathbb E[Y^2mid E_n]=$$$$sum_{i=1}^{n-1}sum_{j=1}^{n-1}mathbb E[X_iX_jmid E_n]=(n-1)(n-2)mathbb E[X_1X_2mid X_1>0,X_2>0]+(n-1)mathbb E[X_1^2mid X_1>0]$$and then variance $mathbb E[Y^2mid E_n]-(mathbb E[Ymid E_n])^2$.
To be found is: $$P(X_1=1mid X_1>0,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)}{P(X_1>0,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)}=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_2+X_3=4)P(X_4=0)}{P(X_1>0,X_2>0,X_3>0,X_1+X_2+X_3=5)P(X_4=0)}=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_2+X_3=4)}{P(X_1>0,X_2>0,X_3>0,X_1+X_2+X_3=5)}=$$$$frac{sum_{s,t>0,s+t=4}P(X_1=1)P(X_2=s)P(X_3=t)}{sum_{r,s,t>0,r+s+t=5}P(X_1=r)P(X_2=s)P(X_3=t)}$$
answered Jan 22 at 9:58
drhabdrhab
102k545136
answered Jan 22 at 9:58
drhabdrhab
102k545136
answered Jan 22 at 9:58
drhabdrhab
102k545136
answered Jan 22 at 9:58
drhabdrhab
102k545136
102k545136
$begingroup$
Thanks for the reply. Since I not good at probability, I hope you can confirm my works. The mean is $E(Y)=(n-1)E(X_i)$ and the variance is $Var(Y)=(n-1)[E(X_i^2)-(E(X_i))^2]$. I don't understand why you write $X_1>0$ as a condition since the value of $X_i$ has already grater than 0.
$endgroup$
– Weihao Huang
Jan 22 at 10:48
$begingroup$
Do you mean that $X_i>0$ (where $i>1$) already implies that $X_1>0$? I do that because I look at this as a process that does not stop (that does not harm, $Y$ can still be defined properly), and work with $X_i$ that are independent. If e.g. conditions like $X_2>0implies X_1>0$ are build in (I avoid that) then this (valuable) independence is lost.
$endgroup$
– drhab
Jan 22 at 11:26
$begingroup$
I see, it not that complicated. $X_i$ is defined as the $|D_1-D_2|$, so it must be greater than $0$.
$endgroup$
– Weihao Huang
Jan 22 at 11:33
$begingroup$
"..so it must be greater than $0$." What do you mean? For every index $i$ it is possible that $D_1^{(i)}=D_2^{(i)}$ or equivalently $X_i=0$.
$endgroup$
– drhab
Jan 22 at 11:38
$begingroup$
I see what you mean, thanks
$endgroup$
– Weihao Huang
Jan 22 at 11:59
add a comment |
$begingroup$
Thanks for the reply. Since I not good at probability, I hope you can confirm my works. The mean is $E(Y)=(n-1)E(X_i)$ and the variance is $Var(Y)=(n-1)[E(X_i^2)-(E(X_i))^2]$. I don't understand why you write $X_1>0$ as a condition since the value of $X_i$ has already grater than 0.
$endgroup$
– Weihao Huang
Jan 22 at 10:48
$begingroup$
Do you mean that $X_i>0$ (where $i>1$) already implies that $X_1>0$? I do that because I look at this as a process that does not stop (that does not harm, $Y$ can still be defined properly), and work with $X_i$ that are independent. If e.g. conditions like $X_2>0implies X_1>0$ are build in (I avoid that) then this (valuable) independence is lost.
$endgroup$
– drhab
Jan 22 at 11:26
$begingroup$
I see, it not that complicated. $X_i$ is defined as the $|D_1-D_2|$, so it must be greater than $0$.
$endgroup$
– Weihao Huang
Jan 22 at 11:33
$begingroup$
"..so it must be greater than $0$." What do you mean? For every index $i$ it is possible that $D_1^{(i)}=D_2^{(i)}$ or equivalently $X_i=0$.
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– drhab
Jan 22 at 11:38
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I see what you mean, thanks
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– Weihao Huang
Jan 22 at 11:59
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Thanks for the reply. Since I not good at probability, I hope you can confirm my works. The mean is $E(Y)=(n-1)E(X_i)$ and the variance is $Var(Y)=(n-1)[E(X_i^2)-(E(X_i))^2]$. I don't understand why you write $X_1>0$ as a condition since the value of $X_i$ has already grater than 0.
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– Weihao Huang
Jan 22 at 10:48
$begingroup$
Thanks for the reply. Since I not good at probability, I hope you can confirm my works. The mean is $E(Y)=(n-1)E(X_i)$ and the variance is $Var(Y)=(n-1)[E(X_i^2)-(E(X_i))^2]$. I don't understand why you write $X_1>0$ as a condition since the value of $X_i$ has already grater than 0.
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– Weihao Huang
Jan 22 at 10:48
$begingroup$
Do you mean that $X_i>0$ (where $i>1$) already implies that $X_1>0$? I do that because I look at this as a process that does not stop (that does not harm, $Y$ can still be defined properly), and work with $X_i$ that are independent. If e.g. conditions like $X_2>0implies X_1>0$ are build in (I avoid that) then this (valuable) independence is lost.
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– drhab
Jan 22 at 11:26
$begingroup$
Do you mean that $X_i>0$ (where $i>1$) already implies that $X_1>0$? I do that because I look at this as a process that does not stop (that does not harm, $Y$ can still be defined properly), and work with $X_i$ that are independent. If e.g. conditions like $X_2>0implies X_1>0$ are build in (I avoid that) then this (valuable) independence is lost.
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– drhab
Jan 22 at 11:26
$begingroup$
I see, it not that complicated. $X_i$ is defined as the $|D_1-D_2|$, so it must be greater than $0$.
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– Weihao Huang
Jan 22 at 11:33
$begingroup$
I see, it not that complicated. $X_i$ is defined as the $|D_1-D_2|$, so it must be greater than $0$.
$endgroup$
– Weihao Huang
Jan 22 at 11:33
$begingroup$
"..so it must be greater than $0$." What do you mean? For every index $i$ it is possible that $D_1^{(i)}=D_2^{(i)}$ or equivalently $X_i=0$.
$endgroup$
– drhab
Jan 22 at 11:38
$begingroup$
"..so it must be greater than $0$." What do you mean? For every index $i$ it is possible that $D_1^{(i)}=D_2^{(i)}$ or equivalently $X_i=0$.
$endgroup$
– drhab
Jan 22 at 11:38
$begingroup$
I see what you mean, thanks
$endgroup$
– Weihao Huang
Jan 22 at 11:59
$begingroup$
I see what you mean, thanks
$endgroup$
– Weihao Huang
Jan 22 at 11:59
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