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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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. – André Nicolas Mar 03 '12 at 15:21
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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 $$\frac{2}{n+1}$$
Henry
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$$ \Pr(x < X_2\ \&\ x<X_3\ \&\ \cdots\ \&\ x<X_n) = \Pr(x<X_2)\cdots\Pr(x<X_n) = \left( e^{-\theta x/2} \right)^{n-1}. $$
Therefore $$ \begin{align} & {} \quad \Pr(X_1 < X_2\ \&\ X_1<X_3\ \&\ \cdots\ \&\ X_1<X_n) \\ \\ & = \mathbb{E}(\Pr(X_1 < X_2\ \&\ X_1<X_3\ \&\ \cdots\ \&\ X_1<X_n \mid X_1)) = \mathbb{E}\left( \left(e^{-\theta X_1/2}\right)^{n-1} \right) \\ \\ & = \int_0^\infty \left( e^{-\theta x/2}\right)^{n-1} \cdot e^{-\theta x} \; \theta\;dx = \int_0^\infty e^{-(n+1)\theta x/2} \; \theta\;dx= \int_0^\infty e^{-(n+1)u} \; 2 \; du = \frac{2}{n+1}. \end{align} $$
(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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1$\theta$ is a scale parameter and so is unlikely to feature in the answer – Henry Mar 03 '12 at 22:24
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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. – Did Apr 04 '12 at 05:57
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