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Calculate $P(X\geq 0, Y \geq -X)$ given $X, Y$ are independent standard normal random variables.

I had thought this could be done as simple as a system of inequalities on $\mathbb{R}^2$ given the entire support integrates to 1, and so the solution is $\frac{3}{8}$.

I was told this isn't a technically correct method, but the solution was correct. Is there some preliminary step that needed made, like regarding conditioning, to make this argument formal?

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    $\begingroup$ It's hard to say if your process is technically correct without knowing the details of your system of inequalities. $\endgroup$ – Brian Tung Dec 15 '19 at 19:57
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    $\begingroup$ Just consider the support of $X,Y\sim N(0,1)$ where they're independent. You can divide R2 into 8 portions, by Y = 0, X = 0, Y = X, Y = -X, and for the above probability you have 3 of these portions. $\endgroup$ – badiki Dec 15 '19 at 20:02
  • $\begingroup$ That's essentially the way I would have done it, so I'm not sure what is lacking either. Maybe rewrite the joint distribution in polar coordinates, making the division into the pizza slices easier? $\endgroup$ – Brian Tung Dec 15 '19 at 20:09
  • $\begingroup$ I would have done the same way. Perhaps you should have explicitly mentioned that you are using the inherent symmetry in because of the Gaussian distributions as this is not true for all distributions. $\endgroup$ – sudeep5221 Dec 15 '19 at 20:11
  • $\begingroup$ @sudeep5221 Maybe the symmetry about the origin with both X and Y was it, I appreciate the feedback! $\endgroup$ – badiki Dec 15 '19 at 20:18
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If you are familiar with this well-known result for the probability of a standard bivariate normal vector falling on the first quadrant, then this question can be answered with the same argument.

It is a fact that whenever $(U,V)$ is bivariate normal with a general mean vector $(\mu_1,\mu_2)$ and a general dispersion matrix $\begin{pmatrix}\sigma_1^2 & \rho\sigma_1\sigma_2\\ \rho\sigma_1\sigma_2 & \sigma_2^2\end{pmatrix}$, then

$$P(U\ge \mu_1,V\ge \mu_2)=P\left(\frac{U-\mu_1}{\sigma_1}\ge 0,\frac{V-\mu_2}{\sigma_2}\ge 0\right)=\frac14+\frac1{2\pi}\arcsin(\rho)$$

Here because $X$ and $Y$ are independent standard normal, $(X,X+Y)$ has a bivariate normal distribution with mean vector $(0,0)$ and dispersion matrix $\begin{pmatrix}1 & 1\\1 & 2\end{pmatrix}$.

Hence, using the result we have

$$P(X\ge 0,X+Y\ge 0)=\frac14+\frac1{2\pi}\arcsin\left(\frac1{\sqrt 2}\right)=\frac38$$

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