I am having difficulty solving the following problem (this is from an old final and there is no solution to check against).
Problem: Let X be a random variable exponentially distributed with parameter λ and let Y = X + 1. Determine the distribution of Y and compute E[Y].
My attempt: Let $F_Y$ be the cumulative distribution function of Y.
$F_Y(y) = P(Y \le y) = P(X \le y - 1) = F_X(y-1)$, which if $(y-1)\ge 0$ is $1 - λe^{-λy}$. By differentiating, $f_Y(y) = λ^2 e^{-λy},$ if $ y \ge1$, $0$ if $y\lt1$. My conclusion is that this is not an exponential distribution because of how the normalizing constant is not the same as the factor in the exponential part, not to mention that the part where the density is nonzero doesn't begin at $y = 0$. But if I am correct about this, then what is the correct answer to this problem?
Edit: actually by integrating from 1 to infinity of the function that I derived, the integral is not even 1 so the function I have is not even a density right?
You did the first step correctly but you wrote pdf instead of the cdf :) Note that $F_X(x) = 1 - e^{-\lambda x}$, so $$ F_Y(y) = F_X(y-1) = 1 - e^{-\lambda (y-1)}, \quad y \ge 1 $$ and therefore $$ f_Y(y) = F'_Y(y) = \lambda e^{-\lambda(y-1)}, \quad y \ge 1. $$
Now $\int_1^\infty f_Y(y) dy = 1$ as expected.
As for the expected value, recall that it is linear, so $$ \mathbb{E}[Y] = \mathbb{E}[X + 1] = \mathbb{E}[X] + 1. $$