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
$\endgroup$ 53 Answers
$\begingroup$$Ee^{-2X}=\int_0^{\infty} e^{-2x} 2 e^{-2x}dx=\frac 1 2 $.
$\endgroup$ 0 $\begingroup$$$\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$$
because it is the integral of a $Exp(4)$ in all its domain
....anyway it i not difficult to calculate it and verify
$\endgroup$ 2 $\begingroup$We are told$$X \sim \text{Exp}(\lambda)$$Therefore,$$\mathrm{E}[X]=\frac{1}{\lambda}$$For justification on this, see [here].() Then, since we are told that $\mathrm{E}[X]=\frac{1}{2}$, we conclude $\lambda=2$. So$$f_X(x)=2e^{-2x}.$$Then, using LOTUS,$$\mathrm{E}[e^{-2X}]=\int_{0}^{\infty}{2e^{-2x}\cdot 2e^{-2x}\mathrm{d}x}=1.$$
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