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In the few sources I've come across which characterize the rate parameter in a Poisson distribution, they described it as a shape parameter. Wouldn't it be more accurate to describe it as a location parameter, since the rate parameter of a Poisson distribution equals the mean, and manipulating the value of the rate parameter determines where on the x-axis the distribution is placed? What have I missed?

Jeff Lowder
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1 Answers1

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Let $X\sim Poisson(\lambda)$.

Yes, $\mathbb{E}\big[X\big] = \lambda$. You are totally correct about that.

But...

$Var\big(X\big)=\lambda$

Therefore, $\lambda$ influences the spread.

When $\lambda$ is small, the distribution has little variability and a mean close to zero, so the distribution is bunched up around $\lambda$ but also has some mass out far, since all positive integers are possible. This means that the distribution is quite skewed.

When $\lambda$ is large, the variability increases along with the mean, allowing the data not to be so bunched up near zero, lessening the skewness. A Poisson distribution with a large $\lambda$ is almost unskewed.

I encourage you to plot some Poisson distributions in a software package like R or Python to get a feel for what’s going on.

Dave
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  • Thanks, Dave. I plotted the PMF for the Poisson using the following values for lambda: 0.5, 5, 10, and 20. All of them seemed to have the same basic shape, but then I noticed that I had some scale compression happening, which had fooled me visually.

    It now seems to me that lambda is functioning as both a scale and a location parameter. Would that be accurate?

     – Jeff Lowder Apr 26 '20 at 20:43
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    Yes, variance is a measure of scale, so $\lambda$ influences both location and scale. Try plotting distributions with much larger $\lambda$ values like $1000$. – Dave Apr 26 '20 at 21:06