Question

Can someone help me to check my answer of moment generating function of Y? it is in my last reply

Let Y be a random variable with probability distribution. P(Y = y) = r*s^(y-1); for y = 1; 2; 3;...; and 0 < s < 1; Derive the value of r that makes this a valid probability distribution?And derive the moment generating function of Y?

Answer 1

my answer is r = 1-s is this right?

Answer 2

Hi @dumbcow, need your help again, you are my life saver!

Answer 3

hmm yes i believe you are right...that would make it a geometric distribution

http://en.wikipedia.org/wiki/Geometric_distribution

Answer 4

Thanks, the next step will be derive the moment generating function of Y. Hopefully I can do it by myself as well

Answer 5

If it is a Prob Distr, it has to fulfill:
\[\sum_{y=1}^{\infty}P(Y=y)=\sum_{y=1}^{\infty}r·s^{y-1}=1\rightarrow r=\frac{ 1 }{ \sum_{y=1}^{\infty}s^{y-1} } =\frac{ 1 }{ \frac{ 1 }{ 1-s}}=1-s\]

Answer 6

Hi @dumbcow, Moment generating function of Y

\[M(t)=E(e^{ty}) = \sum_{y=1}^{\infty} e^{ty} (1-s)s^{y-1} = (1-s) \sum_{y=1}^{\infty} e^{ty} s^{y-1} e^{t} e^{-t}\]
\[= e^{t}(1-s)\sum_{y=1}^{\infty}e^{t(y-1)}s^{y-1} = e^{t}(1-s)\sum_{y=1}^{\infty}(e^{t}s)^{y-1} = \frac{ e^{t}(1-s) }{1-e^{t}s} \]
where \[e^{t}s<1\rightarrow t<-\ln s\]