How can the Poisson distribution be rewritten in terms of P(X <= 1)?

[Bob19]

Hello I'm Presented with the following Poisson distribution question

$$P(X = x) = \frac{e^{-\lambda} \cdot \lambda^{x}}{x!}$$

where $$x \in (1,2,3,\ldots)$$ and $$\lambda > 0$$

Then I'm suppose to show that the above can be re-written if

$$P(X \leq 1) = 1 - e^{- \lambda}$$

Any idears on how I do that?

I'm told $$\sum_{x=0} ^ {\infty} p(x) = \frac{e^{- \lambda} \cdot \lambda^{x}}{x!} = e^{- \lambda} \sum _{x=0} ^{\infty} \frac{\lambda ^{x}}{x!}$$

if $$\lambda ^{x} = 1$$

then $$p(x) = e^{- \lambda}$$

This must give $$P(X \leq 1) = 1- e^{- \lambda}$$

Can anybody please tell me if I'm on the right track here?

Sincerley Bob

 

[HallsofIvy]

Hello I'm Presented with the following Poisson distribution question
$$P(X = x) = \frac{e^{-\lambda} \cdot \lambda^{x}}{x!}$$
where $$x \in (1,2,3,\ldots)$$ and $$\lambda > 0$$
Then I'm suppose to show that the above can be re-written if
$$P(X \leq 1) = 1 - e^{- \lambda}$$
Any idears on how I do that?
I'm told $$\frac{e^{- \lambda} \cdot \lambda^{x}}{x!} = e^{- \lambda} \sum _{x=0} ^{\infty} \frac{\lambda ^{x}}{x!} = e^{- \lambda} e^{\lambda} = 1$$
Then if $$\lambda ^{x} = 1$$

No, you are not told that! Obviously
$$\frac{e^{- \lambda} \cdot \lambda^{x}}{x!}$$
,for specific x, is not the same as sum for all x:
$$e^{- \lambda} \sum _{x=0} ^{\infty} \frac{\lambda ^{x}}{x!}$$
What you are told is that the sum for all x is equal to 1: so that this is a valid probability distribution.
Since x can only take on positive integer values, $P(x \leq 1)$ is exactly the same as $P(x= 1)$ which is
$$\frac{e^{-\lambda}\lambda^1}{1!}= \lambda e^{-\lambda}$$
Assuming, as you say, that that is $1- e^{-\lambda}$, then you are not told that $e^{\lambda}= 1$. You are told, rather, that $e^{\lamba}= 1+ \lambda$ so that
P(X= x) can be written as
$$P(X= x)= \frac{e^{-\lamba}\lamba^x}{x!}= \frac{\lamba^x}{(1+\lambda)x!}$$

 

[Bob19]

Hello Hall and Thank You,

x = 1

then $$\frac{e^{-\lamba}\lamba^1}{1!}= \frac{\lamba^1}{(1+ 1)1!} = e^{- \lambda}$$

Which is then $$P(x \geq 1) = 1- e^{- \lambda}$$

Am I on the right track now ?

Sincerely
Bob

 

[HallsofIvy]

I don't know why the Latex isn't showing properly. What I was trying to say was that IF P(X<= 1)= P(X= 1)= lambda e^-lambda is equal to 1- e^-lambda, then multiplying both sides of the equation by e^lamba we have lambda= e^lambda- 1 or e^lambda= lambda + 1.

Now to "rewrite" P(x)= (lambda^xe^-lambda/x!, just replace that e^-lambda bu 1/(lambda+ 1), getting P(x)= lambda^x/(x!(lambda+ 1)).