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I apologies if I am asking a trivial question, this is something I used to do easily in college now I forgot everything.

I like to know how to estimate the expected value and variance of a mixture of Poisson distribution.

0.3*(((2^x)(e^-2))/x!) + 0.45(((3^x)(e^-3))/x!) + 0.25(((0.5^x)*(e^-0.5))/x!)

enter image description here

Roger V.
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Ahir Bhairav Orai
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  • https://en.wikipedia.org/wiki/Mixture_distribution#Moments – Glen_b Mar 24 '23 at 06:31
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    https://stats.stackexchange.com/questions/16608 gives a thorough answer to finding moments of any mixture where you know the moments of its components. That addresses one possible interpretation of your question. But, since you write "estimate," are you perhaps asking how to estimate the parameters of a Poisson mixture from a sample? Please clarify. – whuber Mar 24 '23 at 13:44

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Suppose $\xi$ is a random variable such that $\mathbb{P}(\xi = k) = 0.3\frac{2^ke^{-2}}{k!} + 0.45\frac{3^ke^{-3}}{k!} + 0.25\frac{\frac{1}{2^k}e^{-\frac{1}{2}}}{k!}$ and $ran(\xi)=\{0,1,2,3,4,5...\}$.

So, by the definition of mathematical expectation you have:

$\mathbb{E}\xi = \sum^{+\infty}_{k=0}k\cdot\mathbb{P}(\xi = k) =\\ \sum^{+\infty}_{k=0}k(0.3\frac{2^ke^{-2}}{k!} + 0.45\frac{3^ke^{-3}}{k!} + 0.25\frac{\frac{1}{2^k}e^{-\frac{1}{2}}}{k!})$

To proceed with this sum you just need to recal, that for any real $z$, $e^z = \sum^{+\infty}_{k=0}\frac{z^k}{k!}$ and we will try to make this sum similar to this expression.

$\sum^{+\infty}_{k=0}k(0.3\frac{2^ke^{-2}}{k!} + 0.45\frac{3^ke^{-3}}{k!} + 0.25\frac{\frac{1}{2^k}e^{-\frac{1}{2}}}{k!}) = \\ \sum^{+\infty}_{k=1}0.3\frac{2^ke^{-2}}{(k-1)!} + \sum^{+\infty}_{k=1}0.45\frac{3^ke^{-3}}{(k-1)!} + \sum^{+\infty}_{k=1}0.25\frac{\frac{1}{2^k}e^{-\frac{1}{2}}}{(k-1)!} = \\= 0.3\cdot e ^{-2}\cdot 2\sum^{+\infty}_{k=1}\frac{2^{k-1}}{(k-1)!} + 0.45\cdot e^{-3} \cdot 3\sum^{+\infty}_{k=1}\frac{3^{k-1}}{(k-1)!} + \\ + \ 0.25\cdot e^{-\frac{1}{2}} \cdot \frac{1}{2}\sum^{+\infty}_{k=1}\frac{\frac{1}{2^{k-1}}}{(k-1)!} = 0.3\cdot e ^{-2}\cdot 2 \cdot e^{2} + 0.45\cdot e^{-3} \cdot \cdot e^{3} \cdot 3 + 0.25\cdot e^{-\frac{1}{2}} \cdot \frac{1}{2} \cdot e^{\frac{1}{2}} = 0.6 + 1.35 + 0.125 = 2.075$.

So, $\mathbb{E}\xi = 2.075$.

To find variance, you just need to use this formula: $\mathbb{V}ar\xi = \mathbb{E}[\xi^2] - (\mathbb{E}\xi)^2$, so you need to find $\mathbb{E}[\xi^2]$.

I hope you now understand how to find $\mathbb{E}[\xi^2]$ but I will you give a trick which is frequently being used to find second moment of poisson random variable.

$\mathbb{E}[\xi^2] = \sum^{+\infty}_{k=0}k^2\cdot\mathbb{P}(\xi = k) =\\ = \sum^{+\infty}_{k=0}(k^2+0)\cdot\mathbb{P}(\xi = k) = \sum^{+\infty}_{k=0}(k^2-k+k)\cdot\mathbb{P}(\xi = k) = \sum^{+\infty}_{k=0}(k^2-k)\cdot\mathbb{P}(\xi = k) + \sum^{+\infty}_{k=0}k\cdot\mathbb{P}(\xi = k) = \sum^{+\infty}_{k=0}(k^2-k)\cdot\mathbb{P}(\xi = k) + \mathbb{E}\xi$.

Now to find the first sum you need to use the same method as I used to find $\mathbb{E}\xi$ and you will be done.

perepelart
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