$X,Y$ are iid from $N(0,1)$. Compute the probability of $X > Y > 0$ [closed]

Question

This question is asked to evaluate $\mathbb{P}(X>Y>0)$. I was thinking if this probability can be evaluated as $\mathbb{P}(X>Y | Y>0) \cdot \mathbb{P}(Y>0)$ by conditional probability. But I had no idea how to calculate $\mathbb{P}(X>Y | Y>0)$.

Any hints will be appreciated!

Answer 1

Think about the joint distribution of $x,y$ as a sphericallly symmetric distribution.

Then consider which region relates to the condition $X>Y>0$ and the question becomes as easy as slicing pizza.

Answer 2

You don't need to bring "conditional probability" concept (the "independence" condition can be utilized to circumvent any conditioning argument) in to solve this question. In addition, the probability $P(X > Y > 0)$ are the same for any continuous distribution $F$ that satisfies $F(0) = \frac{1}{2}$ (we don't even need that $F$ is symmetric about $0$). The case $F = \Phi$ is just a special case.

The proof to the general case goes as follows: \begin{align*} & P(X > Y > 0) \\ =& \int_0^\infty P(X > y)dF(y) \tag{1}\label{1} \\ =& \int_0^\infty (1 - F(y))dF(y) \\ =& \frac{1}{2} - \int_0^\infty F(y)dF(y) \\ =& \frac{1}{2} - \int_{1/2}^1 udu \tag{2}\label{2} \\ =& \frac{1}{2} - \frac{1}{2}\times\left(1 - \frac{1}{4}\right) \\ =& \frac{1}{8}. \end{align*}

Another direction of generalization (i.e., when $X$ and $Y$ are correlated normal r.v.s) can be found in this answer.

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Technical Notes

• $\eqref{1}$ is a direct application of the general theorem: \begin{align*} P[(X, Y) \in B] = \int_{\mathbb{R}^j} P[(x, Y) \in B]dF_X(x) = \int_{\mathbb{R}^k} P[(X, y) \in B]dF_Y(y), \end{align*} where $X$ and $Y$ are independent random vectors with distribution functions $F_X$ and $F_Y$ in $\mathbb{R}^j$ and $\mathbb{R}^k$. When $(X, Y)$ admits a joint density $f(x, y) = f_X(x)f_Y(y)$ with respect to the Lebesgue measure, by drawing a picture you can easily see that \begin{align*} & P(X > Y > 0) = \int_0^\infty\int_y^\infty f(x, y)dx dy \\ =& \int_0^\infty f_Y(y)\int_y^\infty f_X(x)dx dy \\ =& \int_0^\infty P(X > y)dF_Y(y), \end{align*} which coincides with $\eqref{1}$.

• In $\eqref{2}$, I used an intuitive but slack "change of variable" treatment $u = F(y)$. A rigorous proof of showing $\int_0^\infty F(y)dF(y) = \frac{3}{8}$ is more involved, which needs the "integration by parts" formula (cf. Theorem 18.4 in Probability and Measure (3rd edition)). If you are interested in how this works out, you may check this answer for details.

Answer 3

$$ P(X>Y>0) = P(X>Y\mid Y>0)\frac{1}{2} = P(X>Y\mid X,Y>0)\frac{1}{4} = \frac{1}{8}. $$ But, The normal distribution is symmetric so no calculation is needed: The tuple is equally likely to take a value in any of the 8 areas shown below.

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          |   / *******
          |  / Y = X **
          | / *********
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         /|         X
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          |

For non-believers:

$P(X>0) = \int_0^\infty e^{-x^2/2} dx = \frac{1}{2\sqrt{2 \pi}}\int_{-\infty}^\infty e^{-x^2/2} = \frac{1}{2}.$ $P(X>Y\mid X,Y>0) = \frac{2\cdot 2}{(\sqrt{2\pi})^2}\int_0^\infty\int_0^x e^{-0.5(x^2 +y^2)}dydx = \frac{2}{\pi}\int_0^{\infty} \int_0^{\pi/4}r e^{-0.5 r}d\theta dr\\ =\frac{1}{2}\int_0^{\infty} r e^{-0.5 r} dr = \frac{1}{2}(1-0). $

Edit The original formulation asks for $P(X>Y\mid Y>0)$: $\frac{2}{(\sqrt{2\pi})^2}\int_{-\infty}^\infty\int_0^x e^{-0.5(x^2 +y^2)}dydx = \frac{1}{\pi}(0 + \int_0^{\infty} \int_0^{\pi/4}r e^{-0.5 r}d\theta dr)=\frac{1}{4}.$