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Let $\mathcal{F} = \{ X_n, n = 1,2, \ldots\}$ be a family of exponential distributions with parameters $\lambda_n, n=1,2, \ldots$. I am looking for necessary and sufficient conditions for $\mathcal{F}$ to be tight.

Would be grateful, if you could give some ideas or insights. Thanks.

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Do you know Prokhorov's theorem? –  icurays1 Dec 2 '12 at 18:28
    
Hint: Tightness is defined in terms of certain expected values. In this case, you can compute those expected values explicitly. –  Nate Eldredge Dec 2 '12 at 18:31
    
yes, I know Prokhorov's theorem. –  eugen1806 Dec 2 '12 at 18:38
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1 Answer

up vote 2 down vote accepted

If $R>0$, then for each $n$, $$P(|X_n|>R)=\int_R^{+\infty}\lambda_ne^{-\lambda_nx}dx=\int_{\lambda_n R}^{+\infty}e^{-t}dt=e^{-\lambda_nR}.$$ Let $\lambda:=\inf_{n\geqslant 1}\lambda_n$. What can we say if $\lambda>0$?

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Thanks for your comment. In case $\lambda > 0$ we can choose $R > \frac{ln(1/\varepsilon)}{\lambda}$ so that $P\left(|X_n| > R \right) < \varepsilon$. But, in case $\lambda_n \to -\infty$ for example this doesn't work, although there is no guarantee that it is still not possible to choose the compact set $K$ somehow different from $[-R,R]$ so that $P(X_n \in \mathbb \setminus K) < \varepsilon$ if I understand it right. –  eugen1806 Dec 3 '12 at 8:21
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The parameter of an exponential law is positvie. –  Davide Giraudo Dec 3 '12 at 9:51
    
Ah, of course, my bad. So, basically $\inf_{n \geq 1} \lambda_n > 0$ is a sufficient condition for the exponential family to be tight. How could one approach the necessary condition then: i guess it is more difficult since one should claim it independently of how the compact set looks like, i.e. if $\inf_{n \geq 1} \lambda_n = 0$ one should show that there is no way to choose a compact set $K$ of each $\varepsilon$ such that $P(X_n \notin K) < \varepsilon$. –  eugen1806 Dec 3 '12 at 10:18
    
The condition is actually necessary. If $\inf \lambda_n = 0$ there exist a sequence of variables with parameter tending to $0$, and then by prokhorov's theorem, $\mathcal{F}$ is not tight because the dirac in $0$ is not in $\mathcal{F}$. –  saposcat Dec 3 '12 at 15:10
    
Thanks, got it now. –  eugen1806 Dec 4 '12 at 9:02
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