Four accidents occur independently, with each accident following an exponential distribution with a mean of 22.5. What is the expected value of the minimum of the four accidents?
Attempt:
$min(x_1,x_2,x_3,x_4) = n * f(x)* (1-F(x))^{n-1}=4(\frac{1}{22.5}exp(-x/22.5))(exp(-3x/22.5))$
Expectation =$ \int_{0}^{\infty}\frac{4x}{22.5}*exp(-x/5.625)dx = 5.625$
However, I believe this is wrong. Any help would be much appreciated.
In general, let $X_1,\dotsc, X_n$ be iid eponential distribution with mean $1/\lambda$. Then the distribution of the minimum $M$ is $$P(M\leq m) = 1-P(M>m) = 1-(e^{-\lambda m})^{n} = 1-e^{-\lambda nm}.$$ Notice that this shows that $M$ follows an exponential distribution with mean $\frac{1}{\lambda n}$. In our case, $\lambda = 1/22.5, n =4$, and so $$E[M] = \frac{1}{4/22.5} = 5.625.$$ So, our answers agree.
Similarly, if the mean is $90$, then $$E[M] = \frac{1}{4/90} = 22.5$$