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Given a scaled Poisson random variable $Y = cX$, where $X$ is a Poisson random variable with mean $\lambda$, what is the probability mass function (PMF) of $Y$?

The PMF of $X$ is $$p(x) = e^{-\lambda} \lambda^x/x!$$ and MATLAB has a function for it, which is poisspdf. The question is how to compute the PMF of $Y$?

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To avoid confusion, use subscripts to denote the corresponding random variable.

Let $y\in\{0,c,2c,\ldots\}$ and observe that when $c\ne 0$,

$$p_Y(y) = p_X\left(\{x\mid cx=y\}\right) = p_X\left(\{y/c\}\right) = e^{-\lambda}\frac{\lambda^{y/c}}{(y/c)!}.$$

When $c=0$ the only possible value of $y$ is $0$ and $$p_Y(0) = p_X\left(\{x\mid 0x=0\}\right) = p_X\left(\{0,1,2,\ldots\}\right) = 1.$$


The general rule applied here is that when $X$ is any random variable, $f:\mathbb{R}\to\mathbb{R}$ is a measurable function, and $Y=f(X)$, then for any Borel set $\mathcal{B}\subset\mathbb{R}$

$$\Pr(Y\in\mathcal{B}) = \Pr(f(X)\in\mathcal{B}) = \Pr(X\in f^{-1}(B))$$

where

$$f^{-1}(\mathcal B) = \left\{x\in\mathbb{R}\mid f(x)\in\mathcal{B}\right\}.$$

These aren't really facts about probabilities per se: you can see they are merely stating basic properties of functions.

In this particular case the map is $f(x)=cx$ and $\Pr$ is a Poisson distribution; but exactly the same approach works for any (measurable) $f$ and any distribution whatsoever.

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    $\begingroup$ Yes we can! $ $ $\endgroup$
    – amoeba
    Commented Nov 22, 2017 at 21:29
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    $\begingroup$ Carl, please explain why you think not--then we might be able to help you. Exactly what part(s) of this explanation are the cause of your downvote? Incidentally, although I don't know the problem that motivated this question, I can propose some natural applications. One is that you have recorded nonnegative data as counts in bins $0-c$, $c-2c$, $2c-3c$, etc.; you have modeled those counts with a Poisson$(\lambda)$ distribution for the bin indexes $0,1,2,\ldots$; and you now wish to express that model in the original units of measurement by multiplying the bin indexes by $c$. $\endgroup$
    – whuber
    Commented Nov 22, 2017 at 21:39
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    $\begingroup$ @Carl I'm sorry, but somehow we're not communicating. One is free, independently of however the data might behave (which is governed by $\lambda$), to choose whatever bin width one might like. Your statement therefore must be based on some other meaning of what $\lambda$ represents. At some point the English has to give over to the precision of mathematical notation. I have taken some trouble to be careful about that, so I will stand behind it. I would invite you once more to offer a similarly justified and precise solution in your post. $\endgroup$
    – whuber
    Commented Nov 23, 2017 at 0:48
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    $\begingroup$ @Carl I'm sorry about the trouble--it's always tough to respond to downvoting, especially on an old post when you don't have the time (or inclination) to do it immediately. $\endgroup$
    – whuber
    Commented Nov 23, 2017 at 1:02
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    $\begingroup$ Thanks for the clarification @Carl. So what you mean is that scaled Poisson distribution is not a Poisson distribution anymore? Do I understand your statement correctly? If so, we are all in full agreement: of course it's not a Poisson distribution anymore!! Who said otherwise? $\endgroup$
    – amoeba
    Commented Nov 24, 2017 at 8:01

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