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Let $X_1$, $X_2$, · · · , $X_n$ be n independent Exponential random variables with mean 1. Find an expression for E(min($X_1$, $X_2$, · · · , $X_n$ )).

I have been studying random variables for a while but I don't know how to approach this one. Any help is appreciated!

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$P(min(X_1,..,X_n)>t)=P(X_1>t)^{n}=(e^{-t})^{n}=e^{-nt}$. So the density function of $min(X_1,..,X_n)$ is $ne^{-nt}$ (for $t>0$). Can you find the expectation from this?

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Hint

Let $Y=\min(X_1,...,X_n)$.

$$\mathbb P\{Y\geq y\}=\mathbb P\{X_1\geq y,...,X_n\geq y\}=\mathbb P\{X_1\geq y\}^n,$$ where the last equality come from independence. The density function is therefore given by $$f_Y(y)=-n\lambda e^{-\lambda x}(e^{-\lambda x}-1)^{n-1}\cdot \boldsymbol 1_{[0,\infty )}(x).$$

Can you continue from here ?

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