$X$ has a normal distribution. The conditional distribution of another random variable $Y$ given that $X=x$ is a normal distribution with mean $ax+b$ and variance $t^2$, where $a$, $b$, and $t^2$ are constants. How can I prove that the joint distribution of $X$ and $Y$ is bivariate normal distribution?
[Math] Prove that (X,Y) is bivariate normal if X is normal and Y conditionally on X is normal
normal distributionprobability theory
Related Solutions
You have $$ \begin{bmatrix} Z_1 \\ Z_2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ \rho & \sqrt{1-\rho^2} \end{bmatrix} \begin{bmatrix} X \\ Y \end{bmatrix}. $$ The determinant of this matrix is $\sqrt{1-\rho^2}$.
You have the density $$ f_{X,Y}(x,y) = \frac{1}{2\pi} \exp\left( \frac{-1}{2}(x^2+y^2) \right) $$ and $$ \begin{bmatrix} 1 & 0 \\ \rho & \sqrt{1-\rho^2} \end{bmatrix}^{-1} = \begin{bmatrix} 1 & 0 \\ \frac{-\rho}{\sqrt{1-\rho^2}} & \frac{1}{\sqrt{1-\rho^2}} \end{bmatrix} $$ and the determinant of this matrix is $\sqrt{1-\rho^2}$.
That and your assertion about the density will give you the joint density of $W$ and $V$.
If you're looking for the correlation, you can read the covariance and the two variances out of the density function, but that should not be necessary. If you have two random variables $X,Y$ whose covariance matrix is $M$, and you've got $$ \begin{bmatrix} W \\ V \end{bmatrix} = A \begin{bmatrix} X \\ Y \end{bmatrix}, $$ then the covariance matrix of $\begin{bmatrix} W \\ V \end{bmatrix}$ is $$ AMA^T. $$ In this case that is $$ \begin{bmatrix} 1 & 0 \\ \rho & \sqrt{1-\rho^2} \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & \rho \\ 0 & \sqrt{1-\rho^2} \end{bmatrix} = \begin{bmatrix} 1 & \rho \\ \rho & 1 \end{bmatrix}. $$ That gives you $\operatorname{cov}(W,V)$ and the two variances, and since both variances are $1$, the correlation is the covariance.
Nobody is making assumptions in this matter; rather, someone is drawing conclusions. And those conclusions are not based simply on the marginal distributions. Conventionally, one has $$ X_1,\ldots,X_n \overset{\mathrm{i.i.d.}}\sim N(\theta,1), $$ $$ \bar X = \frac{X_1+\cdots+X_n} n \sim N\left(\theta,\frac 1 n \right) $$ This gives use the mean vector and the two variances. Then we have \begin{align} & \operatorname{cov}\left( X_1, \bar X \right) = \operatorname{cov}\left( X_1, \frac{X_1+\cdots+X_n} n \right) \\[6pt] = {} & \frac 1 n \Big( \operatorname{cov}(X_1,X_1)+\operatorname{cov}(X_1,X_2) + \operatorname{cov}(X_1,X_3)+\cdots +\operatorname{cov}(X_1,X_n) \Big) \\[6pt] = {} & \frac 1 n \Big( 1 + 0 + 0 + 0 + \cdots + 0 \Big) = \frac 1 n. \end{align}
Finally, we have the question of why this has a bivariate normal distribution. The definition of the statement that $(U,V)$ has a bivariate normal distribution is that for every pair of constants $a$, $b$ (and "constant" means not random) the random variable $aU+bV$ has a univariate normal distribution. So we need the distribution of $aX_1+b\bar X$. Observe that $$ aX_1+b\bar X = aX_1+b\frac{X_1+X_2+\cdots+X_n} n = \underbrace{\left(a+\frac b n\right)X_1 + b\frac{X_2}n+\cdots+b\frac{X_n}n}. $$ Notice that the part over the underbrace is a sum of independent random variables each of which is normally distributed. Consequently it is itself normally distributed. This conclusion is not based merely on the marginal distributions because it relies on independence.
Best Answer
By definition, $$f_{X,Y}(x,y)=f_X(x)\cdot f_{Y\mid X}(y\mid x)=\frac1{\sqrt{2\pi\sigma^2}}\mathrm e^{-(x-\mu)^2/(2\sigma^2)}\cdot\frac1{\sqrt{2\pi t^2}}\mathrm e^{-(y-ax-b)^2/(2t^2)} $$ Hence the task is to solve for $M$ in $\mathbb R^2$ and $C$ a definite positive $2\times2$ matrix, the fact that for every $z=(x,y)^t$, one has the identity $$ f_{X,Y}(z)=\frac1{2\pi\sqrt{\det C}}\mathrm e^{-\frac12(z-M)^tC^{-1}(z-M)}. $$ Hints: note that the diagonal of $C^{-1}$ is $(1/\sigma^2+a^2/t^2,1/t^2)$ and that $M=(\mu,a\mu+b)^t$. The rest of the proof relies on some simple algebraic manipulations of a degree $2$ polynomial in $(x,y)$.