I have this so far, I\'m now stuck on how to get the distribution function of S.

Question

I have this so far, I'm now stuck on how to get the distribution function of S. And if possible can you check if what I have is the correct probability distribution for S.

X Y Z P(X=x) P(Y=y) P(Z=z) S=X+Y+2Z P(S=s)=P(X=x)P(Y=y)P(Z=z) 0 0 0 1-p 1-p 1-p 0 (1-p)^3 0 0 1 1-p 1-p p 2 (1-p)^2p 0 1 0 1-p p 1-p 1 (1-p)^2p 0 1 1 1-p p p 3 (1-p)^2p 1 0 0 p 1-p 1-p 1 (1-p)^2p 1 0 1 p 1-p p 3 (1-p)p^2 1 1 0 p p 1-p 2 (1-p)p^2 1 1 1 p p p 4 p^3 Let X, YandZ be independent Bernoulli random variables, taking the val- ues 1 and 0 with respective probabilities p and q-1-p. The random variable S X+Y+2Z takes values in the set 10,1, 2, 3, 4). Find the probability distri bution and the distribution function of S

Explanation / Answer

h

hence

Pmf of S
S P(S=s)

0 (1-p)3

1   (1-p)^2p +(1-p)^2p =2(1-p)2p

2 (1-p)^2p +(1-p)p^2 =(1-p)p(p+(1-p))=(1-p)p

3   (1-p)p2+(1-p)p^2=2(1-p)p2

4 p3

now

distribution function

F(S<s) = (1-p)3 S<0

  2(1-p)2p+(1-p)3=(1-p)2(2p+(1-p))=(1-p)2(1+p) S<1

  (1-p)2(1+p) +(1-p)p=(1-p)(1-p2+p) S<2

  2(1-p)p2 +(1-p)(1-p2+p) =(1-p) (1+p2+p) S<3

(1-p) (1+p2+p) +p3 =(1-p+p2-p3+p-p2)+p3=1 S<4

  

X Y Z P(X=x) P(Y=y) P(Z=z) S=X+Y+2Z P(S=s)=P(X=x)P(Y=y)P(Z=z) 0 0 0 1-p 1-p 1-p 0 (1-p)^3 0 0 1 1-p 1-p p 2 (1-p)^2p 0 1 0 1-p p 1-p 1 (1-p)^2p 0 1 1 1-p p p 3 (1-p)p2 1 0 0 p 1-p 1-p 1 (1-p)^2p 1 0 1 p 1-p p 3 (1-p)p^2 1 1 0 p p 1-p 2 (1-p)p^2 1 1 1 p p p 4 p^3

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