Conditional mean and variance with negative random variable

We have the following random variables

$X\sim ber(\frac{1}{3})$

$Y \sim exp(3) \ \text{given} \ \{X=0\}$

$-Y \sim exp(5) \ \text{given} \ \{X=1\}$

Now I have to determine the mean and variance of $Y$

This is my approach, first of all calculate mean and variance of $X$ which is Bernoulli distributed.

$$\mathbb{E}[X]=\frac{1}{3}$$ $$Var(X)=\frac{1}{3}\cdot\frac{2}{3}=\frac{2}{9}$$

Now we move on to the conditional mean and variance

$$Y \sim exp(3) \ \text{given} \ \{X=0\}$$

$$\mathbb{E}[Y|X]=\frac{1}{\lambda}=\frac{1}{3}\cdot X$$

$$\mathbb{E}[Y]=\mathbb{E}[\mathbb{E}[Y|X]]$$ $$\mathbb{E}\left[\frac{1}{3}\cdot X\right]$$ $$\frac{1}{3}\mathbb{E}[X]=\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}$$

$$Var(Y)=\mathbb{E}[Var(Y|X)]+Var(\mathbb{E}[Y|X])$$ $$\mathbb{E}\left[\frac{1}{3}\cdot \frac{2}{3}\cdot X\right]+Var\left(\frac{1}{3}X\right)$$ $$\frac{2}{9}\cdot\mathbb{E}[X]+\frac{1}{9}Var(X)=\frac{2}{9}\cdot \frac{1}{3}+\frac{1}{9}\cdot \frac{2}{9}=\frac{8}{81}$$

But what do I have to do with $-Y$ I don't know how to integrate this random variable into my calculation.

The task is to get $\mathbb{E}[Y]$ and $Var(Y)$

• From $-Y|X=1\sim\text{Exp}(5)$, you should be able to find the distribution of $Y|X=1$. Commented Jan 11, 2018 at 14:01

Hint: The density of $Y$ given $X=1$ is given by $$P(Y=y|X=1) \begin{cases} 0&y>0\\ \lambda e^{\lambda y}&y\leq0 \end{cases}$$ with $\lambda=5$.

$$E[Y]=\sum_xE[Y|X=x]P(X=x) = \frac13 E[Y|X=1]+\frac23E[Y|X=0]\\= \frac13\times(-\frac15)+\frac23\times \frac13$$

Assuming the density of $-Y|X=1$ is of the form $h(t)=5e^{-5t}\mathbf 1_{\{t\ge0\}}$,

you should be able to show that the density of $Y|X=1$ is given by $g(z)=5e^{5z}\mathbf 1_{\{z\le0\}}$.

If you are having trouble finding the mean and variance from the law of total expectation/variance, you can find the distribution of $Y$ directly.

By the total probability theorem, the density of $Y$ is given by

$$f_Y(y)=f_{Y|X=0}(y)\mathrm{Pr}(X=0)+f_{Y|X=1}(y)\mathrm{Pr}(X=1)$$

$$=\frac{5}{3}e^{5y}\mathbf 1_{\{y\le0\}}+2e^{-3y}\mathbf 1_{\{y>0\}}$$

You can find the mean and variance from this density.

You know

\qquad\begin{align}\mathsf P(X{=}0)&=\tfrac 23\\\mathsf P(X{=}1)&=\tfrac 13\\\mathsf E(Y\mid X{=}0)&=\tfrac 13\\\mathsf E(Y\mid X{=}1)&=-\tfrac 15\\\mathsf {Var}(Y\mid X{=}0)&=\tfrac 19\\\mathsf{Var}(Y\mid X{=}1)&=\tfrac 1{25}\end{align}

With these you can use the usual definitions. Or the Laws of Total Expectation and Variance.

\qquad\begin{align}\mathsf E(Y) &= \mathsf E(\mathsf E(Y\mid X)) \\&=\mathsf P(X{=}0)~\mathsf E(Y\mid X{=}0)+ \mathsf P(X{=}1)~\mathsf E(Y\mid X{=}1)\\[3ex]\mathsf{Var}(Y)&=\mathsf E(\mathsf {Var}(Y\mid X))+\mathsf {Var}(\mathsf E(Y\mid X))\\&=\mathsf P(X{=}0)\big(\mathsf{Var}(Y\mid X{=}0)+\mathsf E(Y\mid X{=}0)^2\big)+\mathsf P(X{=}1)\big(\mathsf{Var}(Y\mid X{=}1)+\mathsf E(Y\mid X{=}1)^2\big)-\mathsf E(Y)^2\\&=\mathsf P(X{=}0)\mathsf E(Y^2\mid X{=}0)+\mathsf P(X{=}1)\mathsf E(Y^2\mid X{=}1)-\mathsf E(Y)^2\\&=\mathsf E(Y^2)-\mathsf E(Y)^2\end{align}