For each $n = 1, 2, ....$, suppose that $X_n$ is a continuous random variable with density $$\hspace{10mm}\mathrm{f}(x) = \begin{cases} \frac{1}{2}(1+x)e^{-x}, & \text{if $x \ge 0$ } \\[2ex] 0, & \text{if $x\le 0$} \end{cases}$$ Set $Y_n$=min{$X_1,X_2,....X_n$}. Does n$\cdot$$Y_n$ converges in distribution as n$\to \infty $ ? What will be the limiting distribution of n$\cdot$$Y_n$? Attept: I was tryng to find the distribution of n$\cdot$$Y_n$. But it became too complicated. How should I proceed here. Any help would be much appreciated.

  • $\begingroup$ Yes you are right. Thanks for the edit. $\endgroup$ – RATNODEEP BAIN Aug 17 '18 at 17:25

Let's try to find a distribution function of $nY_n$.\

$P(nY_n \leq t) = 1 - P(min \{ X_1, ..., X_n \} > \frac{t}{n}) = 1 - P(X_1 > \frac{t}{n})\cdot ... \cdot P(X_n > \frac{t}{n})$

We can calculate the tail of the random variable $X_1$ (and all of the variables because I suppose that they are i.i.d).

$P(X_1 > \frac{t}{n}) = \int_{\frac{t}{n}}^{\infty} \frac{1}{2}(1+x) \exp (-x) dx$.

Now, try to do this integral by parts.

  • $\begingroup$ Thanks for the Hint. Now Is the limiting distribution degenerate at t=0? i.e $P(Z=0)=1$ where n$\cdot$$Y_n$ $\to Z$ as $n\to \infty$ $\endgroup$ – RATNODEEP BAIN Aug 17 '18 at 19:32
  • $\begingroup$ We can say that $F(t) = 0$ for $t <= 0$. It suffices main assumption of the cumulative distribution function because $\lim_{t \rightarrow - \infty} F(t) = 0$ For the rest of the $\mathbb{R}$ we have to calculate this integral above. $\endgroup$ – FNTE Aug 17 '18 at 21:17

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