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

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!

• A more general case is discussed in this question. Commented Dec 2, 2023 at 18:08
• Also, think about the type of questions that you can ask here and specifically the topic about self-study questions. And in addition to Zhanxiong's comment, you could do some initial research about previous questions to see whether questions have already been asked before. Asking is a good thing, but make sure that you search yourself as well. Find the right balance. In the case of this website, it is not good if people ask duplicate questions that can be answered with quick research. Commented Dec 2, 2023 at 18:08
• I tried to see if I could search this myselve (those duplicates) and it is not so easy. A problem is that the question is relatively easy and has language that relates with many other related but not identical questions. Commented Dec 2, 2023 at 18:20
• @Sextus Agreed. But I know that precisely this question was asked and answered within the last few years.
– whuber
Commented Dec 3, 2023 at 0:23

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.

• I am sure that this question has already been asked in some way before, but in some way I cannot find it as easy as answering it. Commented Dec 2, 2023 at 18:35

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.

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.

$$P(X>Y>0) = P(X>Y\mid Y>0)\frac{1}{2} = P(X>Y\mid X,Y>0)\frac{1}{4} = \frac{1}{8}.$$ 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.

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



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

• Thanks! But it seems that there exists a little bit mistake in your plot? The inequality $X>Y>0$ should be lower area, not the upper area? Commented Dec 2, 2023 at 22:36