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Let $\{X_n,n>1\}$ be a sequence of i.i.d random variables with the probabilty density function $$f(x) =e^{-(x-a)} \text{ for } x\geq a.$$

Set $Y_n = \min\{X_1,X_2,\ldots,X_n\}$. Show that $Y_n$ converges in probability to $a$ as $n \to \infty$

So there's the question! The only info we have is the definition of the convergence in probability.

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  • $\begingroup$ Write down the definition of convergence in probability. $Y_{n}$ is the minimum of $n$ variables: it is bigger than $y$ only if all $X_{i} > y$. What is the probability that all of them are bigger than $\alpha+\epsilon$, for some $\epsilon > 0$? $\endgroup$
    – megas
    Oct 3, 2016 at 19:59
  • $\begingroup$ P(x>alpha+epsilon) = exp(-epsilon)? @megas $\endgroup$
    – Janono
    Oct 3, 2016 at 20:12

2 Answers 2

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We want to prove that $P(|Y_n-a|>\epsilon)\rightarrow 0$, for any $\epsilon.$

But we have that $P(|Y_n-a|>\epsilon)=P(|X_i-a|>\epsilon, 1\le i\le n)=P(\bigcap_{i=1}^n\{|X_i-a|>\epsilon\})$. Independence now gives that this is:

$P(|X_1-a|>\epsilon)^n=(e^{-\epsilon})^n=e^{-n\epsilon}$. And this converges to 0 as n tends to infinity.

Here I used that $P(|X_1-a|>\epsilon)=\int_{a+\epsilon}^\infty e^{-(x-a)}dx=e^{-\epsilon}$.

PS: This solution assumes that a is positive. (If a is negative f does not represent a probability-density).

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You have $\min > x$ if and only if all of $X_1,\ldots,X_n$ are $>x$.

If $x\ge a$ then $\Pr(X_i>x) = e^{-(x-a)}.$

Since they're independent, the probability that all of them are $>x$ must therefore be $\displaystyle\Big( e^{-(x-a)} \Big)^n$, which is $e^{-n(x-a)}$.

Observe that if $x$ is strictly $>a$, then that approaches $0$ as $n\to\infty$. See if you can work with that

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