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Let $X_1, X_2, . . .$ be a sample from the distribution whose density is:

$f(x) = \begin{cases} \frac{1}{2}(1+x)e^{-x}, & \text{for $x>0$} \\ 0, & \text{otherwise} \end{cases}$

Set $Y_n=min\{X_1,X_2, \dots\}$. Show that $n · Y_n$ converges in distribution as n $\rightarrow 1$, and find the limit distribution.

I did the following:

Since, $F(x)=1-\frac{1}{2}e^{-x}(x+2)$

Then, $P(n \times Y_n\le x)=P(Y_n \le \frac{x}{n})=1-(1-F(x/n))^n=1-\frac{1}{2}e^{-x/n}(2+\frac{x}{n})$

And I don't really see where this function goes once n $\rightarrow \infty$

Any help would be appreciated.

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You have made a mistake in your calculation. What you get is $\lim \{1-[e^{-x/n}(1+\frac x {2n})]^{n}\}=1-e^{-x/2}$ for $x>0$. Hence the limiting distribution is exponential with parameter $\frac 1 2$.

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