What is the Probability Function for a Poisson Distributed Stochastic Variable?

[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:
<br /> P(X=x) = \left\{ \begin{array}{ll}<br /> \frac{{e^{- \lambda}{\lambda ^{x}}}}{{x!}} &amp; \textrm{where} \ x \in (0,1,2,\ldots)&amp;\\<br /> 0 &amp; \textrm{other.}&amp;\\<br /> \end{array} \right.<br />
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

 

[Unco]

G'day, Fred.

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

 

[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

G'day, Fred.

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

 

[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).

 

[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!}} &amp; \textrm{where} \ x \in (-2,-1,0,1) \ \textrm{and} \ \ y \in (0,1,\ldots)&amp;\\<br /> 0 &amp; \textrm{other.}&amp;\\\end{array} \right.

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

Best regards
Fred