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 $X_1,X_2,\ldots,X_n$?
Smallest order statistics
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statistics
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2Hint: 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
2 Answers
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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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0@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 $\frac{2}{n+1}$