# Propability function Question

1. Dec 11, 2005

### Mathman23

Hi Guys
I have Propability function that has caused me some trouble.
X is a stochastic variable which is Poisson distributed with the parameter
$$\lambda > 0$$
The Propability function is therefore:
$$P(X=x) = \left\{ \begin{array}{ll} \frac{{e^{- \lambda}{\lambda ^{x}}}}{{x!}} & \textrm{where} \ x \in (0,1,2,\ldots)&\\ 0 & \textrm{other.}&\\ \end{array} \right.$$
I'm suppose to show
$$P(X \geq 1) = 1 - e^{- \lambda}$$
(step1) I get by inserting into the top formula
$$P(X=1) = \lambda e ^ {- \lambda}$$
My question is how do go from P(X=1) to $$P(X \geq 1)$$ ???
Sincerley
Fred

Last edited: Dec 11, 2005
2. Dec 11, 2005

### Unco

G'day, Fred.

$$P(X \geq 1) = 1 - P(X = 0)$$

3. Dec 11, 2005

### Mathman23

Thanks Mate,

I have second question

If Lambda =1 then $$P(X \leq 2) = \frac{c e^{-1}}{2}$$

where c = 5

Can I show that in a simular way like the first?

Best Regards
/Fred

Last edited: Dec 11, 2005
4. Dec 11, 2005

### Unco

It's similar in that

$$P(X\leq2) = P(X=0) + P(X=1) + P(X=2)$$

Apply the formula for each term and add the fractions (as e-1= 1/e).

5. Dec 11, 2005

### Mathman23

Okay thank You again

This function which now is a to-dimension discrete stochastic vector has the probability function $$p_{X,Y}$$

$$P(X=x,Y=y) = \left\{ \begin{array}{ll}\frac{{c e^{- \lambda}{\lambda ^{x}}}}{{x!}} & \textrm{where} \ x \in (-2,-1,0,1) \ \textrm{and} \ \ y \in (0,1,\ldots)&\\ 0 & \textrm{other.}&\\\end{array} \right.$$

My question is that support $$supp \ p_{X,Y} = (-2,-1,0,1)$$???

Best regards
Fred

Last edited: Dec 11, 2005
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