Conditional variance when there is uncertainity about the distribution of condition
$begingroup$
Assume that $X sim N(0,sigma_x^2)$. Y has the following form
begin{align}
Y &=
begin{cases}
Y_1 sim N(0, sigma_1^2), & text{w.p.} quad mu \
Y_2 sim N(0, sigma_2^2), & text{w.p.} quad (1-mu)
end{cases}
end{align}
where $Cov(X,Y_1) =Sigma_1$, and $Cov(X,Y_2) =Sigma_2$. And also X and Y1, and X and Y2 are both bivariate normally distributed.
How can we calculate $V(X|Y=y)$?
Is the following idea correct?
begin{align*}
V(X|Y=y) = mu V(X|Y=y_1)+(1-mu)V(X|Y=y_2)
end{align*}
Basically, y is realized but it might be in the from of $y_1 in Y_1$ or $y_2 in Y_2$. What do you think?
normal-distribution conditional-expectation conditional-probability variance
asked Jan 21 at 22:15
user229519user229519
105
$endgroup$
add a comment |
$begingroup$
Assume that $X sim N(0,sigma_x^2)$. Y has the following form
begin{align}
Y &=
begin{cases}
Y_1 sim N(0, sigma_1^2), & text{w.p.} quad mu \
Y_2 sim N(0, sigma_2^2), & text{w.p.} quad (1-mu)
end{cases}
end{align}
where $Cov(X,Y_1) =Sigma_1$, and $Cov(X,Y_2) =Sigma_2$. And also X and Y1, and X and Y2 are both bivariate normally distributed.
How can we calculate $V(X|Y=y)$?
Is the following idea correct?
begin{align*}
V(X|Y=y) = mu V(X|Y=y_1)+(1-mu)V(X|Y=y_2)
end{align*}
Basically, y is realized but it might be in the from of $y_1 in Y_1$ or $y_2 in Y_2$. What do you think?
normal-distribution conditional-expectation conditional-probability variance
asked Jan 21 at 22:15
user229519user229519
105
$endgroup$
$begingroup$
I think we need more information to solve this problem: what is the joint distribution of $X$ and $Y_1$ (and $X$ with $Y_2$)? We need to know the distribution of $X$ given $Y = y, Y =Y_1$ (and $Y = y, Y =Y_2$) in order to solve this question. Can we assume that it is a bivariate normal distribution? From the information that you've given so far we can't: this post gives an example where $X, Y$ are both normal and have non-zero covariance but are not bivariate normal stats.stackexchange.com/questions/293818/…
$endgroup$
– Alex
Jan 22 at 8:14
$begingroup$
$X$ and $Y_1$, and $X$ and $Y_2$ are both bivariate normal distribution. But when $y$ is realized it is not known whether it comes from $Y_1$ or $Y_2$. Is this information enough, @Alex.
$endgroup$
– user229519
Jan 22 at 14:14
add a comment |
$begingroup$
Assume that $X sim N(0,sigma_x^2)$. Y has the following form
begin{align}
Y &=
begin{cases}
Y_1 sim N(0, sigma_1^2), & text{w.p.} quad mu \
Y_2 sim N(0, sigma_2^2), & text{w.p.} quad (1-mu)
end{cases}
end{align}
where $Cov(X,Y_1) =Sigma_1$, and $Cov(X,Y_2) =Sigma_2$. And also X and Y1, and X and Y2 are both bivariate normally distributed.
How can we calculate $V(X|Y=y)$?
Is the following idea correct?
begin{align*}
V(X|Y=y) = mu V(X|Y=y_1)+(1-mu)V(X|Y=y_2)
end{align*}
Basically, y is realized but it might be in the from of $y_1 in Y_1$ or $y_2 in Y_2$. What do you think?
normal-distribution conditional-expectation conditional-probability variance
asked Jan 21 at 22:15
user229519user229519
105
$endgroup$
Assume that $X sim N(0,sigma_x^2)$. Y has the following form
begin{align}
Y &=
begin{cases}
Y_1 sim N(0, sigma_1^2), & text{w.p.} quad mu \
Y_2 sim N(0, sigma_2^2), & text{w.p.} quad (1-mu)
end{cases}
end{align}
where $Cov(X,Y_1) =Sigma_1$, and $Cov(X,Y_2) =Sigma_2$. And also X and Y1, and X and Y2 are both bivariate normally distributed.
How can we calculate $V(X|Y=y)$?
Is the following idea correct?
begin{align*}
V(X|Y=y) = mu V(X|Y=y_1)+(1-mu)V(X|Y=y_2)
end{align*}
Basically, y is realized but it might be in the from of $y_1 in Y_1$ or $y_2 in Y_2$. What do you think?
normal-distribution conditional-expectation conditional-probability variance
normal-distribution conditional-expectation conditional-probability variance
asked Jan 21 at 22:15
user229519user229519
105
asked Jan 21 at 22:15
user229519user229519
105
edited Jan 22 at 14:22
user229519
asked Jan 21 at 22:15
user229519user229519
105
asked Jan 21 at 22:15
user229519user229519
105
asked Jan 21 at 22:15
user229519user229519
105
105
$begingroup$
I think we need more information to solve this problem: what is the joint distribution of $X$ and $Y_1$ (and $X$ with $Y_2$)? We need to know the distribution of $X$ given $Y = y, Y =Y_1$ (and $Y = y, Y =Y_2$) in order to solve this question. Can we assume that it is a bivariate normal distribution? From the information that you've given so far we can't: this post gives an example where $X, Y$ are both normal and have non-zero covariance but are not bivariate normal stats.stackexchange.com/questions/293818/…
$endgroup$
– Alex
Jan 22 at 8:14
$begingroup$
$X$ and $Y_1$, and $X$ and $Y_2$ are both bivariate normal distribution. But when $y$ is realized it is not known whether it comes from $Y_1$ or $Y_2$. Is this information enough, @Alex.
$endgroup$
– user229519
Jan 22 at 14:14
add a comment |
$begingroup$
I think we need more information to solve this problem: what is the joint distribution of $X$ and $Y_1$ (and $X$ with $Y_2$)? We need to know the distribution of $X$ given $Y = y, Y =Y_1$ (and $Y = y, Y =Y_2$) in order to solve this question. Can we assume that it is a bivariate normal distribution? From the information that you've given so far we can't: this post gives an example where $X, Y$ are both normal and have non-zero covariance but are not bivariate normal stats.stackexchange.com/questions/293818/…
$endgroup$
– Alex
Jan 22 at 8:14
$begingroup$
$X$ and $Y_1$, and $X$ and $Y_2$ are both bivariate normal distribution. But when $y$ is realized it is not known whether it comes from $Y_1$ or $Y_2$. Is this information enough, @Alex.
$endgroup$
– user229519
Jan 22 at 14:14
$begingroup$
I think we need more information to solve this problem: what is the joint distribution of $X$ and $Y_1$ (and $X$ with $Y_2$)? We need to know the distribution of $X$ given $Y = y, Y =Y_1$ (and $Y = y, Y =Y_2$) in order to solve this question. Can we assume that it is a bivariate normal distribution? From the information that you've given so far we can't: this post gives an example where $X, Y$ are both normal and have non-zero covariance but are not bivariate normal stats.stackexchange.com/questions/293818/…
$endgroup$
– Alex
Jan 22 at 8:14
$begingroup$
I think we need more information to solve this problem: what is the joint distribution of $X$ and $Y_1$ (and $X$ with $Y_2$)? We need to know the distribution of $X$ given $Y = y, Y =Y_1$ (and $Y = y, Y =Y_2$) in order to solve this question. Can we assume that it is a bivariate normal distribution? From the information that you've given so far we can't: this post gives an example where $X, Y$ are both normal and have non-zero covariance but are not bivariate normal stats.stackexchange.com/questions/293818/…
$endgroup$
– Alex
Jan 22 at 8:14
$begingroup$
$X$ and $Y_1$, and $X$ and $Y_2$ are both bivariate normal distribution. But when $y$ is realized it is not known whether it comes from $Y_1$ or $Y_2$. Is this information enough, @Alex.
$endgroup$
– user229519
Jan 22 at 14:14
$begingroup$
$X$ and $Y_1$, and $X$ and $Y_2$ are both bivariate normal distribution. But when $y$ is realized it is not known whether it comes from $Y_1$ or $Y_2$. Is this information enough, @Alex.
$endgroup$
– user229519
Jan 22 at 14:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can use the Conditional Variance Formula:
$$mathbb{V}(X mid Y = y) = mathbb{E}((X - mathbb{E}(Xmid Y = y))^{2}mid Y = y)$$
We need to calculate $mathbb{E}(X mid Y = y)$, and we can do this using a variation of the Law of Total Expectation. Let $A_i$ be the event $Y = Y_i$.
$$mathbb{E}(X mid Y = y) = int xthinspace f_{x mid y}dx = int xfrac{f_{x,y}}{f_y}dx = int xsum_i frac{f_{x,y,A_i}}{f_y}dx$$
$$= int xsum_i frac{f_{x,y,A_i}}{f_{y,A_i}}frac{f_{y,A_i}}{f_{y}}dx= int xsum_i frac{f_{x,y,A_i}}{f_{y,A_i}}frac{f_{y,A_i}}{f_{y}}dx = sum_iint x thinspace f_{x mid y, A_i}mathbb{P}(A_i mid Y = y)dx$$
$$= sum_i int x thinspace f_{x mid y, A_i}dx cdot mathbb{P}(A_i mid Y = y) = sum_i mathbb{E}(X mid Y = y, A_i)mathbb{P}(A_i mid Y = y).$$
So
$$mathbb{E}(X mid Y = y) = mathbb{E}(X mid Y = y, Y = Y_1) mathbb{P}(Y = Y_1 mid Y = y) + mathbb{E}(X mid Y = y, Y = Y_2) mathbb{P}(Y = Y_2 mid Y = y).$$
We can use Bayes' Theorem to calculate the probabilities above:
$$mathbb{P}(Y = Y_1 mid Y = y) = frac{f_{Y_1}(y) mathbb{P}(Y = Y_1)}{f_{Y_1}(y) mathbb{P}(Y = Y_1) + f_{Y_2}(y) mathbb{P}(Y = Y_2)}$$
where
$$f_{Y_i}(y) = frac{1}{sqrt{2pi sigma_i^2}}e^{-y^2/2sigma_i^2} text{for } i = 1, 2.$$
So
$$mathbb{P}(Y = Y_1 mid Y = y) = frac{1/sigma_1 e^{-y^2/2sigma_1^2} mu}{1/sigma_1 e^{-y^2/2sigma_1^2} mu + 1/sigma_2 e^{-y^2/2sigma_2^2} (1 -mu)} = frac{sigma_2 mu e^{-y^2/2sigma_1^2}}{sigma_2 mu e^{-y^2/2sigma_1^2} + sigma_1 (1 - mu)e^{-y^2/2sigma_2^2}}.$$
Also
$$mathbb{P}(Y = Y_2 mid Y = y) = frac{sigma_1 (1 - mu) e^{-y^2/2sigma_2^2}}{sigma_2 mu e^{-y^2/2sigma_1^2} + sigma_1 (1 - mu)e^{-y^2/2sigma_2^2}}.$$
The conditional distribution of $X mid Y = y, Y = Y_i$ is given by
$$X mid Y = y, Y = Y_i sim Nleft(mu_x + rho_ifrac{sigma_x}{sigma_i}(y - mu_i), sigma_x^2 (1 - rho_i^2) right)$$
where $rho_i$ is the correlation of $X$ and $Y_i$, $mu_x = mathbb{E}(X) = 0$ and $mu_i = mathbb{E}(Y_i) = 0$.
Putting this together we have
$$mathbb{E}(X mid Y = y) = rho_1frac{sigma_x}{sigma_1}y cdot p_1 + rho_2frac{sigma_x}{sigma_2}y cdot p_2 = mu_{x mid y},$$
where $p_i = mathbb{P}(Y = Y_i mid Y = y)$.
Returning to the conditional variance formula, we now expand it, conditioning on $Y = Y_i$ again:
$$mathbb{V}(X mid Y = y) = mathbb{E}((X - mu_{xmid y})^{2}mid Y = y)$$
$$= mathbb{E}((X - mu_{xmid y})^{2}mid Y = y, Y = Y_1)mathbb{P}(Y = Y_1 mid Y = y) + mathbb{E}((X - mu_{xmid y})^{2}mid Y = y, Y = Y_2)mathbb{P}(Y = Y_2 mid Y = y).$$
Now $$(X - mu_{xmid y})^{2} = (X - mu_{x mid y,i} + mu_{x mid y,i} - mu_{xmid y})^{2} = (X - mu_{x mid y,i})^2 + 2(X - mu_{x mid y,i})(mu_{x mid y,i} - mu_{xmid y}) + (mu_{x mid y,i} - mu_{xmid y})^2,$$
where $mu_{x mid y,i} = mathbb{E}(X mid Y = y, Y = Y_i)$, so
$$mathbb{E}((X - mu_{xmid y})^{2}mid Y = y = Y_i) = mathbb{E}((X - mu_{x mid y,i})^2 + 2(X - mu_{x mid y,i})(mu_{x mid y,i} - mu_{xmid y}) + (mu_{x mid y,i} - mu_{xmid y})^2mid Y = y = Y_i)$$
$$= sigma_x^2 (1 - rho_i^2) + 0 + (mu_{x mid y,i} - mu_{xmid y})^2.$$
Therefore
$$mathbb{V}(X mid Y = y) = left(sigma_x^2 (1 - rho_1^2) + (mu_{x mid y,1} - mu_{xmid y})^2 right)p_1 + left(sigma_x^2 (1 - rho_2^2) + (mu_{x mid y,2} - mu_{xmid y})^2 right)p_2.$$
$$ = sigma_x^2 + left(- sigma_x^2 rho_1^2 + (mu_{x mid y,1} - mu_{xmid y})^2 right)p_1 + left(- sigma_x^2 rho_2^2 + (mu_{x mid y,2} - mu_{xmid y})^2 right)p_2.$$
answered Jan 23 at 7:45
AlexAlex
674412
$endgroup$
1
$begingroup$
Thanks for your effort, Alex, really appreciate that.
$endgroup$
– user229519
Jan 23 at 20:31
add a comment |
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$begingroup$
You can use the Conditional Variance Formula:
$$mathbb{V}(X mid Y = y) = mathbb{E}((X - mathbb{E}(Xmid Y = y))^{2}mid Y = y)$$
We need to calculate $mathbb{E}(X mid Y = y)$, and we can do this using a variation of the Law of Total Expectation. Let $A_i$ be the event $Y = Y_i$.
$$mathbb{E}(X mid Y = y) = int xthinspace f_{x mid y}dx = int xfrac{f_{x,y}}{f_y}dx = int xsum_i frac{f_{x,y,A_i}}{f_y}dx$$
$$= int xsum_i frac{f_{x,y,A_i}}{f_{y,A_i}}frac{f_{y,A_i}}{f_{y}}dx= int xsum_i frac{f_{x,y,A_i}}{f_{y,A_i}}frac{f_{y,A_i}}{f_{y}}dx = sum_iint x thinspace f_{x mid y, A_i}mathbb{P}(A_i mid Y = y)dx$$
$$= sum_i int x thinspace f_{x mid y, A_i}dx cdot mathbb{P}(A_i mid Y = y) = sum_i mathbb{E}(X mid Y = y, A_i)mathbb{P}(A_i mid Y = y).$$
So
$$mathbb{E}(X mid Y = y) = mathbb{E}(X mid Y = y, Y = Y_1) mathbb{P}(Y = Y_1 mid Y = y) + mathbb{E}(X mid Y = y, Y = Y_2) mathbb{P}(Y = Y_2 mid Y = y).$$
We can use Bayes' Theorem to calculate the probabilities above:
$$mathbb{P}(Y = Y_1 mid Y = y) = frac{f_{Y_1}(y) mathbb{P}(Y = Y_1)}{f_{Y_1}(y) mathbb{P}(Y = Y_1) + f_{Y_2}(y) mathbb{P}(Y = Y_2)}$$
where
$$f_{Y_i}(y) = frac{1}{sqrt{2pi sigma_i^2}}e^{-y^2/2sigma_i^2} text{for } i = 1, 2.$$
So
$$mathbb{P}(Y = Y_1 mid Y = y) = frac{1/sigma_1 e^{-y^2/2sigma_1^2} mu}{1/sigma_1 e^{-y^2/2sigma_1^2} mu + 1/sigma_2 e^{-y^2/2sigma_2^2} (1 -mu)} = frac{sigma_2 mu e^{-y^2/2sigma_1^2}}{sigma_2 mu e^{-y^2/2sigma_1^2} + sigma_1 (1 - mu)e^{-y^2/2sigma_2^2}}.$$
Also
$$mathbb{P}(Y = Y_2 mid Y = y) = frac{sigma_1 (1 - mu) e^{-y^2/2sigma_2^2}}{sigma_2 mu e^{-y^2/2sigma_1^2} + sigma_1 (1 - mu)e^{-y^2/2sigma_2^2}}.$$
The conditional distribution of $X mid Y = y, Y = Y_i$ is given by
$$X mid Y = y, Y = Y_i sim Nleft(mu_x + rho_ifrac{sigma_x}{sigma_i}(y - mu_i), sigma_x^2 (1 - rho_i^2) right)$$
where $rho_i$ is the correlation of $X$ and $Y_i$, $mu_x = mathbb{E}(X) = 0$ and $mu_i = mathbb{E}(Y_i) = 0$.
Putting this together we have
$$mathbb{E}(X mid Y = y) = rho_1frac{sigma_x}{sigma_1}y cdot p_1 + rho_2frac{sigma_x}{sigma_2}y cdot p_2 = mu_{x mid y},$$
where $p_i = mathbb{P}(Y = Y_i mid Y = y)$.
Returning to the conditional variance formula, we now expand it, conditioning on $Y = Y_i$ again:
$$mathbb{V}(X mid Y = y) = mathbb{E}((X - mu_{xmid y})^{2}mid Y = y)$$
$$= mathbb{E}((X - mu_{xmid y})^{2}mid Y = y, Y = Y_1)mathbb{P}(Y = Y_1 mid Y = y) + mathbb{E}((X - mu_{xmid y})^{2}mid Y = y, Y = Y_2)mathbb{P}(Y = Y_2 mid Y = y).$$
Now $$(X - mu_{xmid y})^{2} = (X - mu_{x mid y,i} + mu_{x mid y,i} - mu_{xmid y})^{2} = (X - mu_{x mid y,i})^2 + 2(X - mu_{x mid y,i})(mu_{x mid y,i} - mu_{xmid y}) + (mu_{x mid y,i} - mu_{xmid y})^2,$$
where $mu_{x mid y,i} = mathbb{E}(X mid Y = y, Y = Y_i)$, so
$$mathbb{E}((X - mu_{xmid y})^{2}mid Y = y = Y_i) = mathbb{E}((X - mu_{x mid y,i})^2 + 2(X - mu_{x mid y,i})(mu_{x mid y,i} - mu_{xmid y}) + (mu_{x mid y,i} - mu_{xmid y})^2mid Y = y = Y_i)$$
$$= sigma_x^2 (1 - rho_i^2) + 0 + (mu_{x mid y,i} - mu_{xmid y})^2.$$
Therefore
$$mathbb{V}(X mid Y = y) = left(sigma_x^2 (1 - rho_1^2) + (mu_{x mid y,1} - mu_{xmid y})^2 right)p_1 + left(sigma_x^2 (1 - rho_2^2) + (mu_{x mid y,2} - mu_{xmid y})^2 right)p_2.$$
$$ = sigma_x^2 + left(- sigma_x^2 rho_1^2 + (mu_{x mid y,1} - mu_{xmid y})^2 right)p_1 + left(- sigma_x^2 rho_2^2 + (mu_{x mid y,2} - mu_{xmid y})^2 right)p_2.$$
answered Jan 23 at 7:45
AlexAlex
674412
$endgroup$
1
$begingroup$
Thanks for your effort, Alex, really appreciate that.
$endgroup$
– user229519
Jan 23 at 20:31
add a comment |
$begingroup$
You can use the Conditional Variance Formula:
$$mathbb{V}(X mid Y = y) = mathbb{E}((X - mathbb{E}(Xmid Y = y))^{2}mid Y = y)$$
We need to calculate $mathbb{E}(X mid Y = y)$, and we can do this using a variation of the Law of Total Expectation. Let $A_i$ be the event $Y = Y_i$.
$$mathbb{E}(X mid Y = y) = int xthinspace f_{x mid y}dx = int xfrac{f_{x,y}}{f_y}dx = int xsum_i frac{f_{x,y,A_i}}{f_y}dx$$
$$= int xsum_i frac{f_{x,y,A_i}}{f_{y,A_i}}frac{f_{y,A_i}}{f_{y}}dx= int xsum_i frac{f_{x,y,A_i}}{f_{y,A_i}}frac{f_{y,A_i}}{f_{y}}dx = sum_iint x thinspace f_{x mid y, A_i}mathbb{P}(A_i mid Y = y)dx$$
$$= sum_i int x thinspace f_{x mid y, A_i}dx cdot mathbb{P}(A_i mid Y = y) = sum_i mathbb{E}(X mid Y = y, A_i)mathbb{P}(A_i mid Y = y).$$
So
$$mathbb{E}(X mid Y = y) = mathbb{E}(X mid Y = y, Y = Y_1) mathbb{P}(Y = Y_1 mid Y = y) + mathbb{E}(X mid Y = y, Y = Y_2) mathbb{P}(Y = Y_2 mid Y = y).$$
We can use Bayes' Theorem to calculate the probabilities above:
$$mathbb{P}(Y = Y_1 mid Y = y) = frac{f_{Y_1}(y) mathbb{P}(Y = Y_1)}{f_{Y_1}(y) mathbb{P}(Y = Y_1) + f_{Y_2}(y) mathbb{P}(Y = Y_2)}$$
where
$$f_{Y_i}(y) = frac{1}{sqrt{2pi sigma_i^2}}e^{-y^2/2sigma_i^2} text{for } i = 1, 2.$$
So
$$mathbb{P}(Y = Y_1 mid Y = y) = frac{1/sigma_1 e^{-y^2/2sigma_1^2} mu}{1/sigma_1 e^{-y^2/2sigma_1^2} mu + 1/sigma_2 e^{-y^2/2sigma_2^2} (1 -mu)} = frac{sigma_2 mu e^{-y^2/2sigma_1^2}}{sigma_2 mu e^{-y^2/2sigma_1^2} + sigma_1 (1 - mu)e^{-y^2/2sigma_2^2}}.$$
Also
$$mathbb{P}(Y = Y_2 mid Y = y) = frac{sigma_1 (1 - mu) e^{-y^2/2sigma_2^2}}{sigma_2 mu e^{-y^2/2sigma_1^2} + sigma_1 (1 - mu)e^{-y^2/2sigma_2^2}}.$$
The conditional distribution of $X mid Y = y, Y = Y_i$ is given by
$$X mid Y = y, Y = Y_i sim Nleft(mu_x + rho_ifrac{sigma_x}{sigma_i}(y - mu_i), sigma_x^2 (1 - rho_i^2) right)$$
where $rho_i$ is the correlation of $X$ and $Y_i$, $mu_x = mathbb{E}(X) = 0$ and $mu_i = mathbb{E}(Y_i) = 0$.
Putting this together we have
$$mathbb{E}(X mid Y = y) = rho_1frac{sigma_x}{sigma_1}y cdot p_1 + rho_2frac{sigma_x}{sigma_2}y cdot p_2 = mu_{x mid y},$$
where $p_i = mathbb{P}(Y = Y_i mid Y = y)$.
Returning to the conditional variance formula, we now expand it, conditioning on $Y = Y_i$ again:
$$mathbb{V}(X mid Y = y) = mathbb{E}((X - mu_{xmid y})^{2}mid Y = y)$$
$$= mathbb{E}((X - mu_{xmid y})^{2}mid Y = y, Y = Y_1)mathbb{P}(Y = Y_1 mid Y = y) + mathbb{E}((X - mu_{xmid y})^{2}mid Y = y, Y = Y_2)mathbb{P}(Y = Y_2 mid Y = y).$$
Now $$(X - mu_{xmid y})^{2} = (X - mu_{x mid y,i} + mu_{x mid y,i} - mu_{xmid y})^{2} = (X - mu_{x mid y,i})^2 + 2(X - mu_{x mid y,i})(mu_{x mid y,i} - mu_{xmid y}) + (mu_{x mid y,i} - mu_{xmid y})^2,$$
where $mu_{x mid y,i} = mathbb{E}(X mid Y = y, Y = Y_i)$, so
$$mathbb{E}((X - mu_{xmid y})^{2}mid Y = y = Y_i) = mathbb{E}((X - mu_{x mid y,i})^2 + 2(X - mu_{x mid y,i})(mu_{x mid y,i} - mu_{xmid y}) + (mu_{x mid y,i} - mu_{xmid y})^2mid Y = y = Y_i)$$
$$= sigma_x^2 (1 - rho_i^2) + 0 + (mu_{x mid y,i} - mu_{xmid y})^2.$$
Therefore
$$mathbb{V}(X mid Y = y) = left(sigma_x^2 (1 - rho_1^2) + (mu_{x mid y,1} - mu_{xmid y})^2 right)p_1 + left(sigma_x^2 (1 - rho_2^2) + (mu_{x mid y,2} - mu_{xmid y})^2 right)p_2.$$
$$ = sigma_x^2 + left(- sigma_x^2 rho_1^2 + (mu_{x mid y,1} - mu_{xmid y})^2 right)p_1 + left(- sigma_x^2 rho_2^2 + (mu_{x mid y,2} - mu_{xmid y})^2 right)p_2.$$
answered Jan 23 at 7:45
AlexAlex
674412
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1
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Thanks for your effort, Alex, really appreciate that.
$endgroup$
– user229519
Jan 23 at 20:31
add a comment |
$begingroup$
You can use the Conditional Variance Formula:
$$mathbb{V}(X mid Y = y) = mathbb{E}((X - mathbb{E}(Xmid Y = y))^{2}mid Y = y)$$
We need to calculate $mathbb{E}(X mid Y = y)$, and we can do this using a variation of the Law of Total Expectation. Let $A_i$ be the event $Y = Y_i$.
$$mathbb{E}(X mid Y = y) = int xthinspace f_{x mid y}dx = int xfrac{f_{x,y}}{f_y}dx = int xsum_i frac{f_{x,y,A_i}}{f_y}dx$$
$$= int xsum_i frac{f_{x,y,A_i}}{f_{y,A_i}}frac{f_{y,A_i}}{f_{y}}dx= int xsum_i frac{f_{x,y,A_i}}{f_{y,A_i}}frac{f_{y,A_i}}{f_{y}}dx = sum_iint x thinspace f_{x mid y, A_i}mathbb{P}(A_i mid Y = y)dx$$
$$= sum_i int x thinspace f_{x mid y, A_i}dx cdot mathbb{P}(A_i mid Y = y) = sum_i mathbb{E}(X mid Y = y, A_i)mathbb{P}(A_i mid Y = y).$$
So
$$mathbb{E}(X mid Y = y) = mathbb{E}(X mid Y = y, Y = Y_1) mathbb{P}(Y = Y_1 mid Y = y) + mathbb{E}(X mid Y = y, Y = Y_2) mathbb{P}(Y = Y_2 mid Y = y).$$
We can use Bayes' Theorem to calculate the probabilities above:
$$mathbb{P}(Y = Y_1 mid Y = y) = frac{f_{Y_1}(y) mathbb{P}(Y = Y_1)}{f_{Y_1}(y) mathbb{P}(Y = Y_1) + f_{Y_2}(y) mathbb{P}(Y = Y_2)}$$
where
$$f_{Y_i}(y) = frac{1}{sqrt{2pi sigma_i^2}}e^{-y^2/2sigma_i^2} text{for } i = 1, 2.$$
So
$$mathbb{P}(Y = Y_1 mid Y = y) = frac{1/sigma_1 e^{-y^2/2sigma_1^2} mu}{1/sigma_1 e^{-y^2/2sigma_1^2} mu + 1/sigma_2 e^{-y^2/2sigma_2^2} (1 -mu)} = frac{sigma_2 mu e^{-y^2/2sigma_1^2}}{sigma_2 mu e^{-y^2/2sigma_1^2} + sigma_1 (1 - mu)e^{-y^2/2sigma_2^2}}.$$
Also
$$mathbb{P}(Y = Y_2 mid Y = y) = frac{sigma_1 (1 - mu) e^{-y^2/2sigma_2^2}}{sigma_2 mu e^{-y^2/2sigma_1^2} + sigma_1 (1 - mu)e^{-y^2/2sigma_2^2}}.$$
The conditional distribution of $X mid Y = y, Y = Y_i$ is given by
$$X mid Y = y, Y = Y_i sim Nleft(mu_x + rho_ifrac{sigma_x}{sigma_i}(y - mu_i), sigma_x^2 (1 - rho_i^2) right)$$
where $rho_i$ is the correlation of $X$ and $Y_i$, $mu_x = mathbb{E}(X) = 0$ and $mu_i = mathbb{E}(Y_i) = 0$.
Putting this together we have
$$mathbb{E}(X mid Y = y) = rho_1frac{sigma_x}{sigma_1}y cdot p_1 + rho_2frac{sigma_x}{sigma_2}y cdot p_2 = mu_{x mid y},$$
where $p_i = mathbb{P}(Y = Y_i mid Y = y)$.
Returning to the conditional variance formula, we now expand it, conditioning on $Y = Y_i$ again:
$$mathbb{V}(X mid Y = y) = mathbb{E}((X - mu_{xmid y})^{2}mid Y = y)$$
$$= mathbb{E}((X - mu_{xmid y})^{2}mid Y = y, Y = Y_1)mathbb{P}(Y = Y_1 mid Y = y) + mathbb{E}((X - mu_{xmid y})^{2}mid Y = y, Y = Y_2)mathbb{P}(Y = Y_2 mid Y = y).$$
Now $$(X - mu_{xmid y})^{2} = (X - mu_{x mid y,i} + mu_{x mid y,i} - mu_{xmid y})^{2} = (X - mu_{x mid y,i})^2 + 2(X - mu_{x mid y,i})(mu_{x mid y,i} - mu_{xmid y}) + (mu_{x mid y,i} - mu_{xmid y})^2,$$
where $mu_{x mid y,i} = mathbb{E}(X mid Y = y, Y = Y_i)$, so
$$mathbb{E}((X - mu_{xmid y})^{2}mid Y = y = Y_i) = mathbb{E}((X - mu_{x mid y,i})^2 + 2(X - mu_{x mid y,i})(mu_{x mid y,i} - mu_{xmid y}) + (mu_{x mid y,i} - mu_{xmid y})^2mid Y = y = Y_i)$$
$$= sigma_x^2 (1 - rho_i^2) + 0 + (mu_{x mid y,i} - mu_{xmid y})^2.$$
Therefore
$$mathbb{V}(X mid Y = y) = left(sigma_x^2 (1 - rho_1^2) + (mu_{x mid y,1} - mu_{xmid y})^2 right)p_1 + left(sigma_x^2 (1 - rho_2^2) + (mu_{x mid y,2} - mu_{xmid y})^2 right)p_2.$$
$$ = sigma_x^2 + left(- sigma_x^2 rho_1^2 + (mu_{x mid y,1} - mu_{xmid y})^2 right)p_1 + left(- sigma_x^2 rho_2^2 + (mu_{x mid y,2} - mu_{xmid y})^2 right)p_2.$$
answered Jan 23 at 7:45
AlexAlex
674412
$endgroup$
You can use the Conditional Variance Formula:
$$mathbb{V}(X mid Y = y) = mathbb{E}((X - mathbb{E}(Xmid Y = y))^{2}mid Y = y)$$
We need to calculate $mathbb{E}(X mid Y = y)$, and we can do this using a variation of the Law of Total Expectation. Let $A_i$ be the event $Y = Y_i$.
$$mathbb{E}(X mid Y = y) = int xthinspace f_{x mid y}dx = int xfrac{f_{x,y}}{f_y}dx = int xsum_i frac{f_{x,y,A_i}}{f_y}dx$$
$$= int xsum_i frac{f_{x,y,A_i}}{f_{y,A_i}}frac{f_{y,A_i}}{f_{y}}dx= int xsum_i frac{f_{x,y,A_i}}{f_{y,A_i}}frac{f_{y,A_i}}{f_{y}}dx = sum_iint x thinspace f_{x mid y, A_i}mathbb{P}(A_i mid Y = y)dx$$
$$= sum_i int x thinspace f_{x mid y, A_i}dx cdot mathbb{P}(A_i mid Y = y) = sum_i mathbb{E}(X mid Y = y, A_i)mathbb{P}(A_i mid Y = y).$$
So
$$mathbb{E}(X mid Y = y) = mathbb{E}(X mid Y = y, Y = Y_1) mathbb{P}(Y = Y_1 mid Y = y) + mathbb{E}(X mid Y = y, Y = Y_2) mathbb{P}(Y = Y_2 mid Y = y).$$
We can use Bayes' Theorem to calculate the probabilities above:
$$mathbb{P}(Y = Y_1 mid Y = y) = frac{f_{Y_1}(y) mathbb{P}(Y = Y_1)}{f_{Y_1}(y) mathbb{P}(Y = Y_1) + f_{Y_2}(y) mathbb{P}(Y = Y_2)}$$
where
$$f_{Y_i}(y) = frac{1}{sqrt{2pi sigma_i^2}}e^{-y^2/2sigma_i^2} text{for } i = 1, 2.$$
So
$$mathbb{P}(Y = Y_1 mid Y = y) = frac{1/sigma_1 e^{-y^2/2sigma_1^2} mu}{1/sigma_1 e^{-y^2/2sigma_1^2} mu + 1/sigma_2 e^{-y^2/2sigma_2^2} (1 -mu)} = frac{sigma_2 mu e^{-y^2/2sigma_1^2}}{sigma_2 mu e^{-y^2/2sigma_1^2} + sigma_1 (1 - mu)e^{-y^2/2sigma_2^2}}.$$
Also
$$mathbb{P}(Y = Y_2 mid Y = y) = frac{sigma_1 (1 - mu) e^{-y^2/2sigma_2^2}}{sigma_2 mu e^{-y^2/2sigma_1^2} + sigma_1 (1 - mu)e^{-y^2/2sigma_2^2}}.$$
The conditional distribution of $X mid Y = y, Y = Y_i$ is given by
$$X mid Y = y, Y = Y_i sim Nleft(mu_x + rho_ifrac{sigma_x}{sigma_i}(y - mu_i), sigma_x^2 (1 - rho_i^2) right)$$
where $rho_i$ is the correlation of $X$ and $Y_i$, $mu_x = mathbb{E}(X) = 0$ and $mu_i = mathbb{E}(Y_i) = 0$.
Putting this together we have
$$mathbb{E}(X mid Y = y) = rho_1frac{sigma_x}{sigma_1}y cdot p_1 + rho_2frac{sigma_x}{sigma_2}y cdot p_2 = mu_{x mid y},$$
where $p_i = mathbb{P}(Y = Y_i mid Y = y)$.
Returning to the conditional variance formula, we now expand it, conditioning on $Y = Y_i$ again:
$$mathbb{V}(X mid Y = y) = mathbb{E}((X - mu_{xmid y})^{2}mid Y = y)$$
$$= mathbb{E}((X - mu_{xmid y})^{2}mid Y = y, Y = Y_1)mathbb{P}(Y = Y_1 mid Y = y) + mathbb{E}((X - mu_{xmid y})^{2}mid Y = y, Y = Y_2)mathbb{P}(Y = Y_2 mid Y = y).$$
Now $$(X - mu_{xmid y})^{2} = (X - mu_{x mid y,i} + mu_{x mid y,i} - mu_{xmid y})^{2} = (X - mu_{x mid y,i})^2 + 2(X - mu_{x mid y,i})(mu_{x mid y,i} - mu_{xmid y}) + (mu_{x mid y,i} - mu_{xmid y})^2,$$
where $mu_{x mid y,i} = mathbb{E}(X mid Y = y, Y = Y_i)$, so
$$mathbb{E}((X - mu_{xmid y})^{2}mid Y = y = Y_i) = mathbb{E}((X - mu_{x mid y,i})^2 + 2(X - mu_{x mid y,i})(mu_{x mid y,i} - mu_{xmid y}) + (mu_{x mid y,i} - mu_{xmid y})^2mid Y = y = Y_i)$$
$$= sigma_x^2 (1 - rho_i^2) + 0 + (mu_{x mid y,i} - mu_{xmid y})^2.$$
Therefore
$$mathbb{V}(X mid Y = y) = left(sigma_x^2 (1 - rho_1^2) + (mu_{x mid y,1} - mu_{xmid y})^2 right)p_1 + left(sigma_x^2 (1 - rho_2^2) + (mu_{x mid y,2} - mu_{xmid y})^2 right)p_2.$$
$$ = sigma_x^2 + left(- sigma_x^2 rho_1^2 + (mu_{x mid y,1} - mu_{xmid y})^2 right)p_1 + left(- sigma_x^2 rho_2^2 + (mu_{x mid y,2} - mu_{xmid y})^2 right)p_2.$$
answered Jan 23 at 7:45
AlexAlex
674412
answered Jan 23 at 7:45
AlexAlex
674412
answered Jan 23 at 7:45
AlexAlex
674412
answered Jan 23 at 7:45
AlexAlex
674412
674412
1
$begingroup$
Thanks for your effort, Alex, really appreciate that.
$endgroup$
– user229519
Jan 23 at 20:31
add a comment |
1
$begingroup$
Thanks for your effort, Alex, really appreciate that.
$endgroup$
– user229519
Jan 23 at 20:31
1
1
$begingroup$
Thanks for your effort, Alex, really appreciate that.
$endgroup$
– user229519
Jan 23 at 20:31
$begingroup$
Thanks for your effort, Alex, really appreciate that.
$endgroup$
– user229519
Jan 23 at 20:31
add a comment |
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$begingroup$
I think we need more information to solve this problem: what is the joint distribution of $X$ and $Y_1$ (and $X$ with $Y_2$)? We need to know the distribution of $X$ given $Y = y, Y =Y_1$ (and $Y = y, Y =Y_2$) in order to solve this question. Can we assume that it is a bivariate normal distribution? From the information that you've given so far we can't: this post gives an example where $X, Y$ are both normal and have non-zero covariance but are not bivariate normal stats.stackexchange.com/questions/293818/…
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– Alex
Jan 22 at 8:14
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$X$ and $Y_1$, and $X$ and $Y_2$ are both bivariate normal distribution. But when $y$ is realized it is not known whether it comes from $Y_1$ or $Y_2$. Is this information enough, @Alex.
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– user229519
Jan 22 at 14:14