Let $(X_1,X_2)$ follow bivariate normal distribution with:


$Var(X_1)=1, \ Var(X_2)=2$.



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.stack.imgur.com/Htdva.jpg

I would ask for help to evaluate this double integral

  • $\begingroup$ Hint : express $X_1$ as $\frac{1}{2}Z_2+\frac{\sqrt{3}}{2}Z_3$ with standard normal $Z_2=X_2/\sqrt{2}$ and a standard normal $Z_3$ that is independent of $X_2$ (and of course $Z_2$). Then solve the double integral over $z_2,z_3$ using the product of the $1d$ standard normal densities. You just have to figure out how $Y_1<Y_2^2$ translates into a range for the integration variables $z_2,z_3$. $\endgroup$
    – Kurt G.
    Mar 8, 2022 at 13:29
  • $\begingroup$ $Y_1 < Y_2^2$ is the same as $X_1 < 2X_2$ since $\exp(\cdot)$ is a monotone function. Then $P(X_1 < 2X_2) = \int\limits_{-\infty}^\infty dx_1\int\limits_{\frac{x_1}{2}}^\infty f(x_1, x_2) dx_2,$ where $f$ is the joint-density. $\endgroup$
    – William M.
    Mar 8, 2022 at 15:39
  • $\begingroup$ @WilliamM. According to your instructions ,I have computed the integral which turns out to be like i.stack.imgur.com/Htdva.jpg $\endgroup$
    – Mathtome
    Mar 9, 2022 at 5:13
  • $\begingroup$ Because $E[X_1]=E[X_2]=0$, I think you can use symmetry to argue that $P(X_1 < cX_2) = 1/2$ for any $c$ (regardless of the correlation and marginal variances). $\endgroup$
    – angryavian
    Mar 9, 2022 at 5:36
  • $\begingroup$ @angryavian would you kindly present your argument in the form of an answer..it would be helpful. $\endgroup$
    – Mathtome
    Mar 9, 2022 at 7:01

1 Answer 1


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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