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  1. Let $X_1, X_2$ are independent and both have Poisson distriubution with $\lambda=1$. Find $P\left(\frac{X_1+X_2}{2}\le 1\right)$

  2. Let $X_1, X_2, ...$ are independent and all of them have Poisson distriubution with $\lambda=1$. $\lim_{n\to\infty}P\left(\frac{X_1+...+X_n}{n}\le 1\right)$

My attemption:
1.
$P\left(\frac{X_1+X_2}{2}\le 1\right)=P\left(\frac{X_1+X_2}{2}=0\right)+P\left(\frac{X_1+X_2}{2}=1\right)= P\left(X_1+X_2=0\right)+P\left(X_1+X_2=2\right)$
Now, I don't know why we get assumption about independence. After all, I don't need it. We know, that if $X_1 \tilde {} Poiss(a), X_2\tilde{}Poiss(b) $ then $X_1+X_2\tilde {} Poiss(a+b)$.
So, in our case $X_1+X_2\tilde{} Poiss(2)$, hence $X_1+X_2=X3\tilde{} Poiss(2)$
$P(X_3=0)+P(X_3=1)=\frac{2^0}{0!}e^{-2}+\frac{2^1}{1!}e^{-2}=3e^{-2}$

At this moment I am not going to begin 2. becase I didn't finish 1.. In particular I don't understand why assumption about independence is given.
Can you help me ?

Cm7F7Bb
  • 17,364

1 Answers1

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We need the assumption of the independence. The independence is needed to establish the distribution of the sum. If $X$ and $Y$ are independent random variables with the Poisson distribution, then $X+Y$ has the Poisson distribution too (see here).

However, if $X$ and $Y$ are not necessarily independent, the conclusion no longer holds. Take, for example, $Y=X$. Then $X$ and $Y$ have the Poisson distribution, but $X+Y=2X$ does not have the Poisson distribution anymore. So the assumption of the independence is essential here.

Cm7F7Bb
  • 17,364
  • Thanks for clearing issue with independence. Can you check my solution for 1. and help to solve 2. ? –  Sep 02 '16 at 09:16
  • The cause for $Y=2X$ has no Poisson distribution is (for example) that $P(Y=2X=1) = P(X=1/2) = 0$ . Yeah ? –  Sep 02 '16 at 09:20
  • @HaskellFun That's right. 1. $$P(X_1+X_2\le 2)=P(X_1+X_2=0)+P(X_1+X_2=1)+P(X_1+X_2=2).$$ 2. Not sure about this one. Are you sure that there are no mistakes in the statement? – Cm7F7Bb Sep 02 '16 at 09:30
  • When it comes to 2. - I didn't make a mistake during rewrite task. –  Sep 02 '16 at 09:35
  • So, in case of 1. upper bound is $5e^{-2}$. Yeah ? –  Sep 02 '16 at 09:47
  • Oh, yeah - not bound :) –  Sep 02 '16 at 09:49
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    @HaskellFun It turns out that the second part is not that easy. But the probabilistic sollution given here is really elegant. – Cm7F7Bb Sep 02 '16 at 10:21