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I´m trying to figure out the pdf $f_\min(X_i)$ of $\min(X_i)$, where the distribution of the sample $X_1,...,X_n$ is $\mathcal{E}xp(\lambda)$, where $\lambda$ is the unknown parameter.

I tried with the PDF of the exponential, and it gave me $\lambda*\exp(-\lambda x)$. But I don't know if its correct.

Can you help me out with this?

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The answer can't be correct. We should expect the answer to be a function of $n$.

Let $Y=\min_i X_i$.

Then \begin{align} P(Y \le y) &= 1-P(Y > y)\\ &=1 - \prod_{i=1}^n P\left(X_i > y \right)\\ &=1 - \prod_{i=1}^n \left(1-F_{X_i}(y)\right)\\ &=1 - \left(1-F_{X_1}(y)\right)^n\\ \end{align}

Now, try to differentiate to get the answer.

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  • $\begingroup$ And what would it be in terms of "x" and lambda?. Since I don´t want it to depend on "n". $\endgroup$
    – Ben C.
    Commented Feb 25, 2019 at 7:14
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    $\begingroup$ If something depends on $n$, you can't just ignore it. $\endgroup$
    – Siong Thye Goh
    Commented Feb 25, 2019 at 7:21

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