How Do You Calculate Conditional Probability for a Geometric Distribution?

[Gregg]

The discrete random variable X has probability density P(X=x) =kp^x for x=0,1,... where p \in (0,1). Find normalizing constant k and E(X) as functions of p. For each integer x>0 find P(X>=x) and hence find P(X=y|X>=x) for each integer y>0.

found k=1-p

$$E(X)=\sum kxp^x =p/(1-p)$$

$$P(X>=x) = 1-\sum_{x'=0}^{x} kp^x' = p^{x+1}$$

P(X=y|X>=x) =?

 

[vela]

The definition of the conditional probability P(A|B) is P(A|B) = P(A∩B)/P(B). In this problem, the events are A={y} and B={x, x+1, x+2, ...}. Consider the cases y<x and y≥x separately. Can you take it from there?

 

[Gregg]

looking at it as P(A|B) and the sets A={y}, B{x,x+1,x+2,...}. that's the bit i can't visualise...

is A∩B={y>=x}?

 

[Gregg]

SO is it $$p^{y-x}$$

 

[vela]

looking at it as P(A|B) and the sets A={y}, B{x,x+1,x+2,...}. that's the bit i can't visualise...

is A∩B={y>=x}?

No, they're just normal sets. For instance, if y=5 and x=3, you'd have A={5} and B={3, 4, 5, 6, 7, ...} and their intersection would be A∩B={5}.