Question in probability regarding bivariate normal distribution

probabilityprobability distributionsprobability theoryself-learning

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,
$$(X_1, X_2)\sim \mathscr{N_2}(\mu_1=0, \mu_2=0, \sigma_1^2=1, \sigma_2^2=2,\rho=\frac{1}{2})$$

The general definition of BVN (in fact multivariate normal) is that any linear combination of it's components (here, $X_1$ and $X_2$) are normally distributed (when, pdf exists we can show that this general definition is equivalent to the definition which defines the BVN random vector to be following the BVN pdf, the proof uses characteristic function).

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.,
$$P[Y_1<Y_2^2] = P[e^{X_1}<e^{2X_2}] = P[e^{X_1-2X_2}<1]\\ = P[X_1-2X_2<0]=P[\frac{X_1-2X_2}{\sqrt{9-2\sqrt{2}}}<0]=\Phi(0)=\frac{1}{2}$$

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