Expected value of $e^{-X}$ where $X$ is a random variable

Question

Given $$F_X(x) = \begin{cases} x/2 & 0 < x < 2, \\ 0 & \text{otherwise} \end{cases} $$

determine $E(e^{-X})$.

I have started by deriving the distribution function $f_X(x) = 1/2$

Then I am creating the new random variable $Z = e^{-X}$. $f_Z(x) = e^{-1/2}$.

Then I am solving for $E(Z)$ by taking integral of $xf_Z(x)dx$ from $0$ to $2$.

However this is wrong. Where have I misunderstood the process/erred?

Answer 1

Integration of a density over $\mathbb{R}$ must be $1$, by definition. So you calculation of $f_Z$ is incorrect.

The density of $Z=g(X)$ is not $f_Z(z)=g(f_X(z))$.

What you need to apply is the following $$ E({e^{-X}})=\int_{-\infty}^\infty e^{-x}f_X(x)\,dx. $$

Be careful that you only have $f_X(x)=\frac{1}{2}$ on $(0,2)$.

In general, for any function $h$ (with certain assumptions), one has $$ E(h(X))=\int_{-\infty}^\infty h(x)f_X(x)\,dx. $$

This is also called Law of the unconscious statistician.

Answer 2

If $X$ is between $0$ and $2$ then $Z=e^{-X}$ is between $e^0=1$ and $e^{-2}\approx 0.13533528.$ Therefore $$ \operatorname E(Z) = \int_{e^2}^1 xf_Z(x)\,dx. $$ But it is also equal to $$ \operatorname E(Z) = \int_0^2 e^{-x} f_X(x)\,dx. $$ For that you can integrate by parts.

Your proposed value of $f_Z(x)$ is not correct.

You have \begin{align} F_Z(x) & = \Pr( Z\le x) \\[8pt] & = \Pr(e^{-X} \le x) = \Pr(X\ge - \log_e x) \\[8pt] & = \int_{-\log x}^2 \frac u 2 \, du = \int_2^{-\log x} \frac{-u}2\, du \end{align} and so \begin{align} f_Z(x) & = F_Z'(x) = \frac{\log x} 2 \cdot \frac d {dx} \log x \\[8pt] & = \frac{\log x} 2\cdot \frac 1 x \quad \text{for } e^{-2} < x < 1. \end{align}