I might be missing something very basic, but I have the following scenario. Consider the random variable:
$$ X \sim Exp(1) $$
and call the pdf as $f(x)$. Define a function of this random variable as:
$$ g(X) = \begin{cases} 0 & if \ X=0 \\ 1 & if \ X \in(0,3) \\ 1 + 0.4(X-3) & if \ X\geq3 \end{cases} $$
How can I find $E(g(X))$? I as far as I know there is a the rule for expectations that say
$$ E(g(X)) = \int_{-\infty}^{+\infty} g(x)f(x) dx$$
But I am not sure how to apply this here. I thought about splitting this integral as:
$$ \int_0^0{0f(x)dx} + \int_0^3{1f(x)dx} + \int_3^\infty{\left[1 + 0.4(X-3)\right]f(x)dx}$$
(of course the first one is 0, but I left it there just to be more formal). So my questions are: (1) does this work? (2) is there any shortcut I'm missing, perhaps using some property of the exponential distribution?
Best Answer
This decomposition of the expectation is better understood via indicator functions: write $g(x)$ as $$g(x)=0\mathbb{I}_0(x)+1\mathbb{I}_{(0,3)}(x)+\{1+0.4(x-3)\}\mathbb{I}_{(0,3)^c}(x)$$Then \begin{align} \mathbb{E}[g(X)]&=\mathbb{E}[0\mathbb{I}_0(X)+1\mathbb{I}_{(0,3)}(X)+\{1+0.4(X-3)\}\mathbb{I}_{(0,3)^c}(X)]\\&=\mathbb{E}[0\mathbb{I}_0(X)]+\mathbb{E}[1\mathbb{I}_{(0,3)}(X)]+\mathbb{E}[\{1+0.4(X-3)\}\mathbb{I}_{(0,3)^c}(X)]\\ \end{align} by linearity of the expectation. Now,$$\mathbb{E}[0\mathbb{I}_0(X)]=0$$because the exponential distribution is absolutely continuous wrt Lebesgue measure, hence has no atom (i.e., puts no mass on specific values like $0$). And $$\mathbb{E}[1\mathbb{I}_{(0,3)}(X)]=\int1\mathbb{I}_{(0,3)}(x)\exp\{-x\}\text{d}x=\int_0^3\exp\{-x\}\text{d}x=1-\exp\{-3\}$$while \begin{align}\mathbb{E}[\{1+0.4(x-3)\}\mathbb{I}_{(0,3)^c}(X)]&=\int\{1+0.4(x-3)\}\mathbb{I}_{(0,3)^c}(x)\exp\{-x\}\text{d}x\\&=\int_3^{\infty}\{1+0.4(x-3)\}\exp\{-x\}\text{d}x\\&=\exp\{-3\}+0.4\int_0^\infty x\exp\{-(x+3)\}\text{d}x\\&=\exp\{-3\}+0.4\exp\{-3\}\int_0^\infty x\exp\{-x\}\text{d}x\\&=\exp\{-3\}+0.4\exp\{-3\}\end{align}This leads to$$\mathbb{E}[g(X)]=1+0.4\exp\{-3\}$$