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Let, $T_1,T_2 \cdots T_n $ be i.i.d random variables having reliability function:

$$R-(t) = 1 - \lambda t - o(t)$ as $t \rightarrow 0$$.

Show that $X_n = n \min(T_1,T_2 \cdots T_n )$ has assymtoticaly an exponential distribution as $n \rightarrow \infty$

Note: $F(t) = 1 - R(t)$ , $F(t)$ being the cdf.

What I have done so far is:

1.) Calculated the cdf of $\min(T_1,T_2 \cdots T_n )$, which is: $[ 1 - {R(t)}^n ]$

2.) Put this value and obtain the cdf of $X_n$, which results in: n $[ 1 - {(1 - \lambda t - o(t))}^n ]$

After that I am lost.

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    $\begingroup$ As usual with minima, one should try to estimate $P(X_n>x)$ for some fixed $x$, when $n\to\infty$. $\endgroup$ – Did Jul 16 '15 at 12:08
  • $\begingroup$ @Did I have added what I have done. p.s. please comment $\endgroup$ – Croma14 Jul 16 '15 at 12:20
  • $\begingroup$ Your point 2) is careless and wrong: If the CDF of $X_n/n$ is $1-(1-R(t))^n$, the CDF of $X_n$ is... $\endgroup$ – Did Jul 16 '15 at 12:52
  • $\begingroup$ @Did Its should be... $1/n *$ cdf of $ X_n/n$ $\endgroup$ – Croma14 Jul 16 '15 at 13:06
  • $\begingroup$ "Its should be... 1/n∗cdfofXn/n" What? No, really no. $\endgroup$ – Did Jul 16 '15 at 13:07
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$$P\{\min\{T_1,\dots,T_n \}\le t \}=P\{X_n/n\le t\}=1-[R(t)]^n$$

Then

$$P\{X_n \le t \}=P\{X_n/n \le t/n \}=1-[R(t/n)]^n\rightarrow 1-e^{-\lambda t}$$

as $n\rightarrow \infty$ by plugging the expansion of $R(t/n)$ around $0$.

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