Show that the Poisson Mass Function has the same value for λ and λ-1

[Dakarai]

Homework Statement

Show that, quite generally, the Poisson Mass Function has the same value at λ (i.e. the average) and at (λ-1).

Homework Equations

The Poisson Mass Function is
p(x) = [e^(-λ) * λ^(x)]/(x!)

The Attempt at a Solution

I started out by plugging in λ-1 into the equation to get p(x)= [e^(-λ-1) * (λ-1)^(x)]/(x!) but now I've gotten stuck. I'm not sure how to progress and prove this other than plug in different x and λ values. I've seen the graphs and such that indicate that λ and λ-1 share the same value, but I just can't figure out how to mathematically prove it.

 

[Charles Link]

If I'm not mistaken, ## \lambda ## does not change in what you are trying to show. One thing you can do is let ## p(x-1)=p(x) ## and solve for ## x ##.

 

[Dakarai]

Thank you for the advice. I'm very confused as to how this would actually show that p(x|λ) = p(x|λ-1).

 

[Charles Link]

This one is quite simple I think. ## e ^ {-\lambda} * \lambda^ {(x-1)}/(x-1)!=e ^ {-\lambda} * \lambda^x/x! ## . A little algebra and you have the answer. (Note x is an integer here).

 

[Dakarai]

Oh, I see where I'm confused. I'm not looking to solve for x-1, but rather λ-1. x remains x for this equation.

 

[Dakarai]

If I'm not mistaken, ## \lambda ## does not change in what you are trying to show. One thing you can do is let ## p(x-1)=p(x) ## and solve for ## x ##.

I reread this, and so I've attached the homework question (and the mentioned graph) to better help describe what I'm getting at. I'm very lost how to prove this.http://imgur.com/NyuuHmT

 

[Dakarai]

 

[Charles Link]

## \lambda ## is a parameter that is pre-selected that gives the shape of the curve. For a given ## \lambda ## there will be two consecutive integers, x-1 and x that have the same value for ## p ## i.e. ## p(x-1)=p(x) ##. That equation, in post #4 is easy to solve. ## x!/(x-1)!=x ##. I could show you the rest, but the Forum rules want the OP to do some of the work. When you get this answer for ## x ## it tells you where this "peak" occurs, i.e. for a given pre-selected ## \lambda ##, where the ## p ## function levels off with two equal values before it drops back down. If ## p(x) ## were a continuous function, you would set ## dp(x)/dx=0 ## to find this peak. Instead, ## x ## has only integer values, so you set ## p ## equal for two consecutive integers.

 

[Ray Vickson]

Thank you for the advice. I'm very confused as to how this would actually show that p(x|λ) = p(x|λ-1).

No: look at the diagram you supplied. The parameter ##\lambda## is fixed, and you are being asked to show that ##p(x|\lambda) = p(x-1|\lambda)## when ##x = \lambda##. That's all.