Let $(X_1,X_2)$ follow bivariate normal distribution with:
$\mathbb{E}(X_1)=\mathbb{E}(X_2)=0$
$Var(X_1)=1, \ Var(X_2)=2$.
AND
$\text{Corr}(X_1,X_2)=\frac{1}{2}$
Let $Y_i=e^{X_i}, \ i=1,2$.
Calculate $\mathbb{P}(Y_1 < {Y_2}^2)$
In this problem, I was looking up a bit on the joint pdf of bivariate normal but don't find a way. It would be helpful to get an insightful answer explaining the same.
Edit: I have calculated the integral which turns out to be like:[The integral computed ][1]
How to calculate this integral.
[1]: https://i.sstatic.net/Htdva.jpg
I would ask for help to evaluate this double integral
Best Answer
We are given the Bivariate Normal(BVN) random vector,
$$(X_1, X_2)\sim \mathscr{N_2}(\mu_1=0, \mu_2=0, \sigma_1^2=1, \sigma_2^2=2,\rho=\frac{1}{2})$$
If we use the above fact then,
$$X_1-2X_2 \sim \mathscr{N}(0,\ 9-2\sqrt{2})$$
(How? Hint: Use linearity of expectation and the formula for variance of linear combination of r.v.s)
So, now returning the the question asked by O.P.,
$$P[Y_1<Y_2^2] = P[e^{X_1}<e^{2X_2}] = P[e^{X_1-2X_2}<1]\\ = P[X_1-2X_2<0]=P[\frac{X_1-2X_2}{\sqrt{9-2\sqrt{2}}}<0]=\Phi(0)=\frac{1}{2}$$