Probability question, given a Random Variable $X\sim \exp (\lambda )$, let $Y=g(X)$, what is $E(Y)$?

Question

How would I solve this?

Let $X\sim \exp (\lambda )$ be a random variable. Furthermore, let $Y=g(X)$, with$$g(X)=\left \{\begin{matrix}a & \text{if} & X<a, \\ X & \text{if} & X>a.\end{matrix}\right .$$

What is $\mathbb{E}(Y)$?

The correct answer is $\mathbb{E}(Y)=a+\dfrac{1}{\lambda}e^{-\lambda a}$.

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I tried many times, but I could never get to the correct answer. Any help is appreciated!

Answer 1

By definition:

$$E[g(x)]=\int_0^\infty g(x)f_X(x)~dx$$

where $f_X(x)=\lambda e^{-\lambda x}$, so we have:

$$E[g(x)]=\int_0^a af_X(x)~dx+\int_a^\infty xf_X(x)~dx$$

$$E[g(x)]=\int_0^a a\lambda e^{-\lambda x}~dx+\int_a^\infty x\lambda e^{-\lambda x}~dx$$

You can procced from here to compute these two integrals, and you will get that answer.

Answer 2

Just use the memoryless property.

$\qquad\begin{align}\mathsf E(Y) &= \mathsf E(a\,\mathbf 1_{X<a})+\mathsf E(X\,\mathbf 1_{X\geq a}) &&\text{Partitioning} \\[1ex]&= a\,\mathsf P(X<a)+ \mathsf P(a\leq X)\,\mathsf E(X\mid a\leq X) &&\mathsf E(Z\mid X\in A) =\dfrac{\mathsf E(Z\,\mathbf 1_{X\in A})}{\mathsf P(X\in A)} \\[1ex]&= &&\text{Exponential Distributions are Memoryless} \\[1ex]&= &&X\sim\mathcal{Exp}(\lambda) \\[1ex] &= a +\lambda^{-1}\mathrm e^{-\lambda a}&&\text{Algebraic Cancelations}\end{align}$