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Suppose $X_1$ has exponential distribution with mean $\frac{1}{\theta}$ and $X_2,\ldots,X_n$ have exponential distribution with mean $\frac{2}{\theta}$. also suppose $X_1,X_2,\ldots,X_n$ are independent.how likely that $X_1$ be smallest order statistics in sample $X_1,X_2,\ldots,X_n$?

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    Hint: The *min* of exponentials $X_2$ up to $X_n$ with parameter $\lambda$ (so mean $1/\lambda$) has exponential distribution parameter $(n-1)\lambda$. Thus you will be comparing two exponentials.2012-03-03

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$$ \Pr(x < X_2\ \&\ x

Therefore $$ \begin{align} & {} \quad \Pr(X_1 < X_2\ \&\ X_1

(I've just revised this after Henry pointed out an obvious flaw that I had neglected. I shall return shortly to check the details.)

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    $\theta$ is a scale parameter and so is unlikely to feature in the answer2012-03-03
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    @Henry : I see your point. I'll do some further edits.....2012-03-04
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    You exchanged the parameters of $X_1$ on the one hand and of all the other $X_i$s on the other hand.2012-04-04
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    @Didier : OK, I've done some revisions.2012-04-07
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Taking André Nicolas's comment, replace $X_1$ by the lower of independent $Y$ and $Z$ each with exponential distribution with mean $\frac{2}{\theta}$. The minimum of $Y$ and $Z$ has an exponential distribution with mean $\frac{1}{\theta}$, the same distribution as $X_1$.

So now you are asking what is the probability that either $Y$ or $Z$ is the lowest of $Y,Z, X_2,\ldots, X_n$, where all are continuous iid. By symmetry this is $$\frac{2}{n+1}$$