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Let $X_1, \ldots , X_n$ be iid random variables with common pdf

$f(x) = e^{-(x-\theta)}$ when $x>\theta$ and $f(x)=0$ elsewhere. Here $\theta$ is a fixed parameter.

This pdf is called the shifted exponential. Let $Y_n = \min\{X_1, \ldots ,X_n\}$. Prove that $Y_n \rightarrow \theta$ in probability, by first obtaining the cdf of $Y_n$.

My answer:

The cdf is . . . $F_{Y_n}(x) = -e^{-x+\theta}$ when $x>\theta$ and $F_{Y_n}(x) = 0$ elsewhere

$$P[|Y_n - \theta| < \epsilon] = P[\theta - \epsilon < Y_n < \theta + \epsilon] = F_{Y_n}(\theta + \epsilon) - F_{Y_n}(\theta - \epsilon) = -e^{-\epsilon}$$

But now I have to show that $-e^{-\epsilon}$ approaches 1 as $n$ approaches infinity, right? But I'm confused, because there is no $n$, so it just stays the same no matter what $n$ is...

Thanks in advance

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Can you explain how you arrived at the cdf for $Y_n$? As Alex mentioned below, the cdf you mention is not quite right. In fact, $F_{Y_n}$ consists $n$ in the power of the exponent. – Srikanth Jan 27 '13 at 23:06
up vote 1 down vote accepted

When you write the cdf of $Y_n$, that is your mistake. You are interested in the minimum of a sequence of random variables. Think about what it means for $P(Y_n\leq x)$, that means that there is at least one $X_i$ for $1\leq i\leq n$ with $X_i\leq x$. Alternatively, $P(Y_n\geq x)$ means $X_i\geq x$ for each $i=1,2,\ldots,n$. The $X_i$ are independent so...

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So the probability that $Y_n$ is greater than some value is the same as the probability that all the Xi's are greater. So instead of using $-e^{-x+\theta}$, we just use $1+e^{-n(x+\theta)}$ since they are iid, right? – user58289 Jan 28 '13 at 9:32
So The answer is $1+e^{-n(2\theta + \epsilon)} \rightarrow 1$, right? – user58289 Jan 28 '13 at 9:33

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