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Suppose X and Y are independent normal r.v. $N(0,1)$. An exercise about calculating conditional expectation on events is:

(1) Find $\mathbb{E}[X+Y|X \ge 0, Y \ge 0]$.

(2) Find the probability density function of $X+Y$ given $X\ge 0, Y \ge 0$.

I am not familiar with the definition of conditional expectation on events (I only have learnt the definition on $\sigma$-field). Here are my attempts. I wonder how to calculate (2)(some of my attempt is as follows) and whether my solution of (1) is correct:

(1): By symmetricity: \begin{align} \mathbb{E}[X + Y|X\ge 0,Y \ge 0] &= 2\mathbb{E}[X|X\ge 0, Y\ge 0] \\ &= 2\mathbb{E} [X\frac{f(X,\chi_{X\ge 0},\chi_{Y\ge 0})}{f(\chi_{X \ge 0},\chi_{Y\ge 0})}] \\ &= 4\mathbb{E} \int_{\mathbb{R}^+} x \frac{1}{\sqrt{2\pi}}e^{-x^2/2} dx \\ &= 2\sqrt{\frac{2}{\pi}} \end{align} (2) By basic property of independent normal distribution, $A = X+Y \sim N(0,2), B = X-Y \sim N(0,2)$ are independent. (Details: (X,Y) is 2d Gaussian vector, hence the independence equals to the uncorrectedness.) Hence I try to transform the problem from form $X,Y$ to $A,B$.

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Pina Rinith
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1 Answers1

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For (1), your notation is unclear. The second equality should read: $$ 2E(X\mid X\ge0, Y\ge0)=\frac{2E(XI_{(X\ge0, Y\ge0)})}{P(X\ge0,Y\ge0)},\tag1$$ using the formula for expectation of a random variable conditional on an event: $$E(Z\mid B)=\frac{E(ZI_B)}{P(B)}$$ where $I_B$ $(=\chi_B)$ denotes the indicator of event $B$. Now simplify the numerator in (1) using independence: $$E(XI_{(X\ge 0)}I_{(Y\ge0)})=E(XI_{(X\ge 0)})E(I_{(Y\ge0)})=E(XI_{(X\ge0)}) P(Y\ge0)=\frac12 E(XI_{(X\ge0)})$$. Therefore the RHS of (1) equals $$\frac{E(XI_{(X\ge0)})}{P(X\ge0, Y\ge0)}=4\int_0^\infty x\,\frac1{\sqrt{2\pi}}e^{-x^2/2}dx=2\sqrt{\frac2\pi}. $$

For (2), use the hint in this answer to derive the conditional cumulative distribution function $P(X+Y\le z\mid X\ge0, Y\ge0)$, and differentiate with respect to $z$ to obtain the conditional probability density. The answer will be $$2\sqrt2\left[\Phi\left(\frac z{\sqrt 2}\right)-\Phi\left(- \frac z{\sqrt 2}\right)\right]\phi\left(\frac z{\sqrt 2}\right),$$ where $\phi$ denotes the standard normal density function and $\Phi$ the standard normal cumulative distribution function.

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