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We know that if $X_1$ and $X_2$ are independent random variables such that $ X_1 \sim \text{Poisson}(\lambda_1) $ and $X_2 \sim \text{Poisson}(\lambda_2)$ that $X_1+X_2 \sim \text{Poisson}(\lambda_1+\lambda_2)$

Is there any result about a linear combination of two independent poisson random variables $a_{1} X_1+a_2 X_2$ where $a_1, a_2 \in \mathcal{R}$?

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    $\begingroup$ This question has been asked before (though I cannot find it) and the answer is No: not much can be said, even in the case when $a_1$ and $a_2$ are restricted to be integers rather than arbitrary numbers in $\mathbb R$. $\endgroup$ Commented Feb 8, 2015 at 14:34
  • $\begingroup$ I agree but my main aim is to decompose a Poisson process into a linear decomposition of two Poisson processes that are independent that is why I want to know if there are any clues or it is impossible. $\endgroup$
    – user144209
    Commented Feb 8, 2015 at 14:38
  • $\begingroup$ @user144209 Have you looked at infinite divisibility of Poisson processes? $\endgroup$
    – Clement C.
    Commented Feb 8, 2015 at 14:39
  • $\begingroup$ yes, but the infinite divisibility of Poisson process is related to the special case where $a_1=a_2$ what if $a_1 \neq a_2$. $\endgroup$
    – user144209
    Commented Feb 8, 2015 at 14:43
  • $\begingroup$ If you want to decompose a Poisson process of rate $\lambda$ into two independent Poisson processes of rates $\lambda_1 > 0$ and $\lambda_2 = \lambda-\lambda_1 > 0$, create a sequence of iid Bernoulli random variables $B_i$ with parameter $p = \frac{\lambda_1}{\lambda}$ and assign the $i$-th arrival in the base process to sub-process #1 or #2 according as $B_i = 1$ or $B_i = 0$. Note: the sequence of Bernoulli random variables needs to be independent of the Poisson process too. $\endgroup$ Commented Feb 8, 2015 at 14:43

2 Answers 2

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If $X \sim Poisson(\lambda)$ and $Y=cX$, then $\mathbf{E}Y = c \lambda \neq \mathbf{Var} Y = c^2 \lambda$.

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  • $\begingroup$ What if he included the restriction $\frac{a_1-a_1^2}{a_2-a_2^2}=\frac{E[X_2]}{E[X_1]}$? (This restriction came from setting $E[a_1X_1+a_2X_2]=var(a_1X_1+a_2X_2)$) $\endgroup$
    – user103828
    Commented Feb 8, 2015 at 14:36
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$$\mathbb{E}e^{t(a_1X_1+a_2X_2)}=\frac{1}{1-\frac{t}{\lambda/a_1}}\frac{1}{1-\frac{t}{\lambda/a_2}}=\frac{1}{1-a_1/a_2}\frac{1}{1-\frac{t}{\lambda/a_1}}+\frac{1}{1-a_2/a_1}\frac{1}{1-\frac{t}{\lambda/a_2}}$$Taking the inverse Laplace transform you'll find the pdf of $a_1X_1+a_2X_2$.

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