2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Yes you are right. Thanks for the edit. $\endgroup$ – RATNODEEP BAIN Aug 17 '18 at 17:25
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.