Suppose $X$ is a continuous random variable with PDF: $$\begin{cases} e^{-(x-c)}\ \ \text{when }x > c \\ 0\ \quad \quad\text{when}\ x \leq c \end{cases}$$
a. Find $\mathbb{E}(X)$
b. Find $\mathbb{E}(X-c)$
c. Find $\mathbb{E}(X/c)$
I know how to calculate expected value over a given interval $[a,b]$: $$\int_a^b xf(x) \mathrm{d}x.$$ But the lack of interval as well as the $c$ is throwing me off. I'm thinking the interval should be from $0$ to infinity, but that's as much as I've got.
On
In general, the Linearity of Expectation says that for any constants $a,b$ and random variable $X$, then: $$\mathsf E(aX+b) ~=~ a~\mathsf E(X)+b$$
If you can find $\mathsf E(X)$, then you can find $\mathsf E(X-c)$ and $\mathsf E(\tfrac 1 c X)$
$$\begin{align} \mathsf E(X)~ &=~ \int_c^\infty xe^{-(x-c)}\operatorname d x \\&=~ \int_0^\infty (y+c)e^{-y}\operatorname d y \\&= ~1+c \end{align}$$
a.
$E(X)=\int_{c}^{\infty}xe^{-(x-c)}dx=\int_{c}^{\infty}xe^{-x+c}dx=-\left(x+1\right)\mathrm{e}^{c-x}\vert_{c}^{\infty}=0-(-(x+1))=x+1$
b.
$E(X-c)=E(X)-c=x+1-c$
c.
$E(X/c)=E(X)/c=\frac{x+1}{c}$
Hint:
For the integral part $\int_{c}^{\infty}xe^{c-x}dx$, you could use integral by parts, and let $f(x)=x$ and $g'(x)=e^{c-x}$. Then $f'(x)=1$ and $g(x)=-e^{c-x}$. Then you just have to integrate $-xe^{c-x}-\int_{c}^{\infty}-e^{c-x}dx$.