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There is a sequence of independent random Poisson variables $Y_1, Y_2,\ldots$ with divergent parameter $Y_i \sim \text{Poisson}(\frac{1}{n})$. Does this sequence converge almost surely? In $L^2$ space? In probability?

I am really confused as I did some research on my own and almost every question I come up to has to do with the sum of a sequence; but I do not believe this has to do anything with the sum of $Y_n$ right? to prove it converges almost surely, I have to prove that $Y_n$ approaches the limit of $e^{-1}*\sum_{i=1}^{\infty} 1/i$? Any hints would be appreciated. Thanks.

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    $\begingroup$ Why is the parameter $\frac1n$ divergent? $\endgroup$
    – zoli
    Apr 1, 2018 at 9:41

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Since $Y_n\geqslant 0$ a.s. we have $$ \|Y_n\|_2 = \mathbb E[Y_n^2] = \mathsf{Var}(Y_n) + \mathbb E[Y_n^2] = \frac1n + \frac1n^2\stackrel{n\to\infty}\longrightarrow0, $$ so that $Y_n\to 0$ in $L^2$. This implies that $Y_n\to 0$ in probability. Moreover, $$ \sum_{n=1}^\infty \mathbb P(Y_n\geqslant 1) = \sum_{n=1}^\infty \left(1-e^{-\frac1n}\right) = +\infty, $$ so from the second Borel-Cantelli lemma we conclude that $$ \mathbb P\left(\limsup_{n\to\infty} \{Y_n\geqslant 1\}\right) = 1, $$ and hence $Y_n$ does not converge to zero almost surely.

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