Show that $P(X_2 \leq x \mid X_1 = x)\to0$ when $x\to-\infty$, where $(X_1,X_2)$ is normal with known conditional distribution

Question

Assume that that the conditional distribution of $X_2$ conditionally on $X_1$ is $$f_{X_2\mid X_1}(x_2\mid x_1) = \frac{f_{X_1,X_2}(x_1,x_2)}{f_{X_1}(x_1)} = \frac{1}{\sqrt{ 2 \pi (1- \rho^{2})}} \exp \left( -\frac{(x_{2}-\rho x_{1})^{2}}{{2(1 - \rho^{2})}} \right) $$ Show that $$\lim\limits_{x \rightarrow -\infty} P(X_2 \leq x \mid X_1 = x) = 0$$

What I have done: $$ \lim\limits_{x \rightarrow -\infty}{ P(X_{2} \leq x \mid X_{1} = x)} = \lim\limits_{x \rightarrow -\infty}{ \int_{-\infty}^{x}} \frac{1}{ \sqrt{2 \pi (1- \rho^{2})}} \exp \left({- \frac{(x_{2}-\rho x)^{2}}{{2(1 - \rho^{2})}}} \right) dx_{2} $$

Now I can see that this has to be $0$ since $x\to-\infty$ and I integrate from $-\infty$ to $-\infty$. Is this correct? Any hint is welcome.

Answer 1

I have to show that $\lim\limits_{x \rightarrow -\infty}{ P(X_{2} \leq x | X_{1} = x)} = 0$ $$ \lim\limits_{x \rightarrow -\infty}{ P(X_{2} \leq x | X_{1} = x)} = \lim\limits_{x \rightarrow -\infty}{ \int_{-\infty}^{x}} \frac{1}{ \sqrt{2 \pi(1- \rho^{2})}} e^{- \frac{(x_{2}-\rho x)^{2}}{{2(1 - \rho^{2})}}} dx_{2} $$ Substituting $$ z = x_{2}-\rho x$$ $$ dz = dx_{2} $$ and changing the bounds $$ z = -\infty - \rho x = -\infty$$ $$ z= x - \rho x = x(1- \rho) $$ yields $$\lim\limits_{x \rightarrow -\infty}{ \int_{-\infty}^{(1-\rho)x}} \frac{1}{ \sqrt{2 \pi(1- \rho^{2})}} e^{- \frac{(z)^{2}}{{2(1 - \rho^{2})}}} dz = 0 $$

Answer 2

I may provide some more statistical (maybe even intuitive) then mathematical answer to this question. The formula for conditional normal value distribution is $X_{1}|X_{2}=x\sim(\mu_{1}+\frac{\sigma_{1}}{\sigma_{2}}\rho(x-\mu_{2}),(1-\rho^2)\sigma_{1}^2)$. So we may see that variance is indendend of $x$ so the expectation only matters. Consequently as $x\to-\infty$ expectation, by absolute value, becomes very large. Then if $\frac{\sigma_{1}}{\sigma_{2}}\rho\ne0$ and $\frac{\sigma_{1}}{\sigma_{2}}\rho\ne1$ the difference between expectation and $x$ becomes higher and higher (untill very huge), so $x$ appears somewhere very far in the left side of the tail (because variance remains constant) that makes probability of getting values less then $x$ approach zero. If $\frac{\sigma_{1}}{\sigma_{2}}\rho=0$ then $X_{1}$ and $X_{2}$ are independend so we deal with marginal normal distribution where $P(X_{1}\leq-\infty)=0$. Finally if $\frac{\sigma_{1}}{\sigma_{2}}\rho=1$ then the difference between expectation and $x$ always will be $\mu_{1}-\mu_{2}$ so in this case it seems to me that the limit does not hold.