Assume that $X\sim exp(0.001)$.
We are talking about cars and $X$ is the lifetime of a car (at days).
We buy two cars, what is $E(X)$ and $Var(X)$?
Can I use the fact that: $E(X)=\frac1\lambda$ and $Var(X)=\frac1{\lambda^2}$?
So: $E(X)=\frac{1}{0.001}\Rightarrow$ $E(2X)=2E(X)=2000$?
And same thing for $Var(X)$ - $Var(2X)=4Var(X)=\frac{4}{0.001^2}$.
I'm right? Or I miss something...
Thank you!
Let $X_1$ and $X_2$ denote the lifetimes of two cars. We assume that the $X_i$ are i.i.d. exponential: they are independent because one car breaking down does not affect the other car, and they are both exponential with parameter $\lambda = 0.001$, hence identically distributed.
The length of time you have two cars is the length of time until the first car breaks down, i.e. $Y = \min\{X_1, X_2\}$. Note that \begin{align*} P(Y > y) &= P(X_1 > y, X_2 > y) \\ &= P(X_1 > y)P(X_2 > y) \\ &= P(X_1 > y)^2\\ &= (e^{-\lambda y})^2, \end{align*} from which we see that $Y \sim \exp(2\lambda)$ (since $P(Y \leq y) = 1 - e^{-2\lambda y}$, which is the cdf of $\exp(2\lambda)$). Therefore, $E Y = \frac{1}{0.002}$ and $\text{Var}(Y) = \frac{1}{0.002^2}.$
Note that this is a different question from asking what $E(X_1 + X_2)$ is (i.e. the total amount of time you have any car).