91影视

Y is a normal random variable with mean 渭 and variance ${蟽}^2$

The conditional distribution of X, given that Y = y

Mean y and variance 1

Let $Z$is a standard normal random variable and is independent of $Y$. The conditional distribution of $Y+Z$given that $Y=y$is normal with mean $y$and variance $1$. Therefore, the joint distribution of $X,Y$ is the same that of $Y+Z,Y$.

Therefore, the joint distributior of $X,Y$ is the same that of $Y+Z,Y$.

Y is a normal random variable with mean 渭 and variance ${蟽}^2$

The conditional distribution of X, given that Y = y

Mean y and variance 1

The pair of the random variables $Y+Z,Y$has a bivariate normal distribution, because $Y+Z$and $Y$are both linear combinations of the independent normal random variables $Y$ and $Z$.

The pair of the random variables $Y+Z,Y$has a bivariate normal distribution

Y is a normal random variable with mean 渭 and variance ${蟽}^2$

The conditional distribution of X, given that Y = y

Mean y and variance 1

Find $E\left(X\right),\mathrm{Var}鈦�\left(X\right)$, and $蚁$

Now the calculation is,

$Y~N\left(渭,{蟽}^2\right)\mathrm{\&}Z~N(0,1)$

$E\left(X\right)=E(Y+Z)$

$=E\left(Y\right)+E\left(Z\right)$

$=渭+0$

$=渭$

Then,

$\mathrm{Var}鈦�\left(X\right)=\mathrm{Var}鈦�(Y+Z)$

$=\mathrm{Var}鈦�\left(Y\right)+\mathrm{Var}鈦�\left(Z\right)+2\mathrm{Cov}鈦�(Y,Z)$

$={蟽}^2+1+2\left(0\right)$

$=1+{蟽}^2$

Calculate the correlation coefficient $蚁$in below;

$\mathrm{Corr}鈦�(X,Y)=\mathrm{Corr}鈦�(Y+Z,Y)$

$=\frac{\mathrm{Cov}鈦�(Y+Z,Y)}{\sqrt{\mathrm{Var}鈦�(Y+Z)}\sqrt{\mathrm{Var}鈦�\left(Y\right)}}$

$=\frac{E\left[\right(Y+Z\left)Y\right]鈭�E(Y+Z)E\left(Y\right)}{\sqrt{\mathrm{Var}鈦�\left(X\right)}\sqrt{\mathrm{Var}鈦�\left(Y\right)}}$

$=\frac{E\left[Y^2\right]鈭�E\left(YZ\right)鈭�E\left(Y\right)E\left(Y\right)鈭�E\left(Z\right)E\left(Y\right)}{\sqrt{\mathrm{Var}鈦�\left(X\right)}\sqrt{\mathrm{Var}鈦�\left(Y\right)}}$

$=\frac{E\left[Y^2\right]鈭�\left[E\right(Y){]}^2鈭�\{E\left(YZ\right)鈭�E\left(Z\right)E\left(Y\right)\}}{\sqrt{\mathrm{Var}鈦�\left(X\right)}\sqrt{\mathrm{Var}鈦�\left(Y\right)}}$

$=\frac{\mathrm{Var}鈦�\left(Y\right)鈭�\mathrm{Cov}鈦�(Y,Z)}{\sqrt{\mathrm{Var}鈦�\left(X\right)}\sqrt{\mathrm{Var}鈦�\left(Y\right)}}$

$=\frac{{蟽}^2鈭�0}{\sqrt{{蟽}^2+1}脳\sqrt{{蟽}^2}}$

$Y$and $Z$are independent

$=\frac{蟽}{\sqrt{{蟽}^2+1}}$

Therefore it has been found that$E\left(X\right)=渭,\mathrm{Var}鈦�\left(X\right)=1+{蟽}^2,$and$\mathrm{Corr}鈦�(X,Y)=\frac{蟽}{\sqrt{{蟽}^2+1}}$.

Y is a normal random variable with mean 渭 and variance ${蟽}^{2_{}}$

The conditional distribution of X, given that Y = y

Mean y and variance 1

We need to calculate the mean of the conditional distribution of $Y$given $X=x$.

$E[Y鈭�X=x]=E\left(Y\right)+\mathrm{Corr}鈦�(X,Y)\frac{\sqrt{\mathrm{Var}鈦�\left(Y\right)}}{\sqrt{\mathrm{Var}鈦�\left(X\right)}}(x鈭�E(X\left)\right)$

$=渭+\frac{蟽}{\sqrt{{蟽}^2+1}}\frac{\sqrt{{蟽}^2}}{\sqrt{1+{蟽}^2}}(x鈭�渭)$

$=渭+\frac{{蟽}^2}{\left({蟽}^2+1\right)}(x鈭�渭)$

Therefore it has been found that$E[Y鈭�X=x]=渭+\frac{{蟽}^2}{\left({蟽}^2+1\right)}(x鈭�渭)$.

Y is a normal random variable with mean 渭 and variance ${蟽}^2$

The conditional distribution of X, given that Y = y

Mean y and variance 1

From part (d), the mean of the conditional distribution of $Y$given $X=x$is

$E[Y鈭�X=x]=渭+\frac{{蟽}^2}{\left({蟽}^2+1\right)}(x鈭�渭)$

We need to calculate that the variance of the conditional distribution of $Y$given $X=x$.

$\mathrm{Var}鈦�[Y鈭�X=x]=\mathrm{Var}鈦�\left(Y\right)\left\{1鈭�[\mathrm{Corr}鈦�(X,Y){]}^2\right\}$

$={蟽}^2\left\{1鈭�{\left[\frac{蟽}{\sqrt{{蟽}^2+1}}\right]}^2\right\}$

$={蟽}^2\left\{1鈭�\frac{{蟽}^2}{{蟽}^2+1}\right\}$

$={蟽}^2\left\{\frac{1}{{蟽}^2+1}\right\}$

$=\frac{{蟽}^2}{1+{蟽}^2}$

Therefore, the conditional distribution of $Y$ given $X=x$ is normal with mean$E[Y鈭�X=x]=渭+\frac{{蟽}^2}{\left({蟽}^2+1\right)}(x-渭)$ and variance$Var[Y鈭�X=x]=\frac{{蟽}^2}{{蟽}^2+1}$.