MHB Probability function (p.f) of a random variable

Click For Summary
To find the probability function of the Bernoulli random variable W derived from the Poisson random variable T, the relationship between the two variables is established. The probability that T equals zero is given by P(T=0) = (λ^0/0!)e^(-λ). Consequently, the probability that W equals one is equal to P(W=1) = P(T=0), while the probability that W equals zero is P(W=0) = 1 - P(T=0). Therefore, the probability function for W can be expressed as P(W=1) = e^(-λ) and P(W=0) = 1 - e^(-λ). This provides a clear method to determine the probability function for the random variable W.
serbskak
Messages
1
Reaction score
0
If one has a Bernoulli random variable W that is derived from a Variable T (Poisson λ), by the following rules W = (if T=0 then W=1 and if T>0 then W=0), I am having trouble finding the pf for W. Any suggestions about how to proceed forward?
 
Physics news on Phys.org
Welcome to the forum.

The distribution for $T$ says that $P(T=0)=\frac{\lambda^0}{0!}e^{-\lambda}$. By assumption $P(W=1)=P(T=0)$ and $P(W=0)=1-P(T=0)$.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
4K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K