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

1. Apr 11, 2016

### Dakarai

1. The problem statement, all variables and given/known data
Show that, quite generally, the Poisson Mass Function has the same value at λ (i.e. the average) and at (λ-1).

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

3. 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.

2. Apr 11, 2016

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 $x$.

3. Apr 11, 2016

### 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).

4. Apr 11, 2016

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).

5. Apr 11, 2016

### 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.

6. Apr 12, 2016

### Dakarai

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

7. Apr 12, 2016

### Dakarai

8. Apr 12, 2016

$\lambda$ 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.

Last edited: Apr 12, 2016
9. Apr 12, 2016

### Ray Vickson

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.