Expected value of e^-2x

Question

I have the following problem:

The expectation of an exponentially distributed random variable $X$ is equals to $1/2$. Compute $\mathbb{E}[e^{-2x}]$.

The final answer is: $1/2$

I already know that $\mathbb{E}[x]$ of an exponential distribution equals $1/λ$. So $λ=2$.

That makes the probability density function: $f(x) = 2e^{-2x}$

I also know that $\mathbb{E}[e^{-2x}]$ equals the integral of $e^{2x} \cdot f(x)$. But when I solve this equation I get $(-1/2)\cdot e^{-4x}$.

Can I get feedback to get the final solution?

Ter

Answer 1

We are told $$X \sim \text{Exp}(\lambda)$$ Therefore, $$\mathrm{E}[X]=\frac{1}{\lambda}$$ For justification on this, see [here].(https://en.wikipedia.org/wiki/Exponential_distribution) Then, since we are told that $\mathrm{E}[X]=\frac{1}{2}$, we conclude $\lambda=2$. So $$f_X(x)=2e^{-2x}.$$ Then, using LOTUS, $$\mathrm{E}[e^{-2X}]=\int_{0}^{\infty}{2e^{-2x}\cdot 2e^{-2x}\mathrm{d}x}=1.$$

Answer 2

$Ee^{-2X}=\int_0^{\infty} e^{-2x} 2 e^{-2x}dx=\frac 1 2 $.

Answer 3

$$\mathbb{E}[e^{-2x}]=\int_0^{+\infty}2 e^{-4x}dx=\frac{1}{2}\int_0^{+\infty}4 e^{-4x}dx=\frac{1}{2}$$

the integral

$$\int_0^{+\infty}4 e^{-4x}dx=1$$

because it is the integral of a $Exp(4)$ in all its domain

....anyway it i not difficult to calculate it and verify