$E[\Pr(X<Y)]$ where $X$ and $Y$ are normally distributed?

Question

Sorry if this is badly worded, not sure of the best way to express this.

Suppose we have $X \sim N(0,1)$ and $Y \sim N(0,\sigma)$ and the CDF of $X: F(z) = \Pr(X < z)$. If we take the expectation of $F(X)$ we get 0.5, as discussed here:

Expectation of CDF of continuous random variable $X$, evaluated at $X$

But what if we take the expectation of $F(Y)$? In this case the PDF in question is no longer the derivative of the CDF, so the integral seems less obvious. Any help would be appreciated!

Answer 1

This actually works for any distributions that are symmetric about $0$, i.e. $X$ and $-X$ have the same distribution and $Y$ and $-Y$ have the same distribution, where the distribution of $X$ is continuous.

Then the CDF of $X$ satisfies $$F(y) = \mathbb P(X \le y) = \mathbb P(X \ge -y) = 1 - F(-y)$$ so $$ \mathbb E[F(Y)] = 1 - \mathbb E[F(-Y)] = 1 - \mathbb E[F(Y)]$$ and therefore $\mathbb E[F(Y)] = 1/2$.

Answer 2

Use this expression for expectation, $\mathbb{E}(X)= \int_{\mathbb{R}} xdF(x)$ $$\mathbb{E}(F(Y))=\int_{\mathbb{R}} F(y) dF(y) = \int_0^1 x dx = \frac{1}{2}$$

Answer 3

To be clear, we’d better to distinguish two notations, $\Pr \left[ X<Y\left| Y \right. \right]$ and $\Pr \left[ X<Y \right]$. The former means the conditional probability of the event $X<Y$ given the value of $Y$. The latter means the probability of the same event where $Y$ is treated as a random variable. It is essential to understand the relation between the probability and expectation of an event $A$, that is,$\Pr \left[ A \right]=\mathbb{E}\left[ {{I}_{\left\{ A \right\}}} \right]$ where ${{I}_{\left\{ A \right\}}}$ is a indication function for an event A.

Saying that, if you mean $\mathbb{E}\left[ \Pr \left[ X<Y \right] \right]$ as the expectation of the latter notation $\mathbb{E}\left[ \Pr \left[ X<Y \right] \right]=\Pr \left[ X<Y \right]$ since $\Pr \left[ X<Y \right]$ is a non-random variable and the expectation of a non-random variable is itself. If you mean $\mathbb{E}\left[ \Pr \left[ X<Y \right] \right]$ as the expectation of the former notation, that is, $\underset{Y}{\mathop{\mathbb{E}}}\,\left[ \Pr \left[ X<Y\left| Y \right. \right] \right]$, we can calculate as follows.

$$\underset{Y}{\mathop{\mathbb{E}}}\,\left[ \Pr \left[ X<Y\left| Y \right. \right] \right]=\underset{Y}{\mathop{\mathbb{E}}}\,\left[ \underset{X}{\mathop{\mathbb{E}}}\,\left[ {{I}_{\left\{ X<Y \right\}}}\left| Y \right. \right] \right]=\mathbb{E}\left[ {{I}_{\left\{ X<Y \right\}}} \right]=\Pr \left[ X<Y \right].$$

After all, $\mathbb{E}\left[ \Pr \left[ X<Y \right] \right]=\Pr \left[ X<Y \right]$ in the both cases.

Since $X\sim N\left( {{\mu }_{X}},\sigma _{X}^{2} \right)$ and $Y\sim N\left( {{\mu }_{Y}},\sigma _{Y}^{2} \right)$,$$X-Y\sim N\left( {{\mu }_{X}}-{{\mu }_{Y}},\sigma _{Z}^{2} \right),where\,\sigma _{Z}^{2}:=\sigma _{X}^{2}+\sigma _{Y}^{2}-\operatorname{cov}\left( X,Y \right).$$

We get $$\Pr \left[ X<Y \right]=\Pr \left[ X-Y<0 \right]=\Pr \left[ {\left( X-Y-\left( {{\mu }_{X}}-{{\mu }_{Y}} \right) \right)}/{{{\sigma }_{Z}}}\;<{-\left( {{\mu }_{X}}-{{\mu }_{Y}} \right)}/{{{\sigma }_{Z}}}\; \right]=\Phi \left( {-\left( {{\mu }_{X}}-{{\mu }_{Y}} \right)}/{{{\sigma }_{Z}}}\; \right)$$ where $\Phi \left( {} \right)$ is the C.D.F of the standard normal distribution.

Finally we get $\mathbb{E}\left[ \Pr \left[ X<Y \right] \right]=\Phi \left( {-\left( {{\mu }_{X}}-{{\mu }_{Y}} \right)}/{{{\sigma }_{Z}}}\; \right)$. When ${{\mu }_{X}}={{\mu }_{Y}}=0$ as your case, $\mathbb{E}\left[ \Pr \left[ X<Y \right] \right]=\Phi \left( 0 \right)={1}/{2}\;.$

Furthermore, even if neither $X$ nor $Y$ are normal distributions, $\mathbb{E}\left[ \Pr \left[ X<Y \right] \right]={1}/{2}\;$ holds only if $X-Y$ is a symmetrical distribution with 0 mean.