Respuesta :
The question is incomplete. The complete question is :
Let X be a random variable with probability mass function
P(X =1) =1/2, P(X=2)=1/3, P(X=5)=1/6
(a) Find a function g such that E[g(X)]=1/3 ln(2) + 1/6 ln(5). You answer should give at least the values g(k) for all possible values of k of X, but you can also specify g on a larger set if possible.
(b) Let t be some real number. Find a function g such that E[g(X)] =1/2 e^t + 2/3 e^(2t) + 5/6 e^(5t)
Solution :
Given :
[tex]$P(X=1)=\frac{1}{2}, P(X=2)=\frac{1}{3}, P(X=5)=\frac{1}{6}$[/tex]
a). We know :
[tex]$E[g(x)] = \sum g(x)p(x)$[/tex]
So, [tex]$g(1).P(X=1) + g(2).P(X=2)+g(5).P(X=5) = \frac{1}{3} \ln (2) + \frac{1}{6} \ln(5)$[/tex]
[tex]$g(1).\frac{1}{2} + g(2).\frac{1}{3}+g(5).\frac{1}{6} = \frac{1}{3} \ln (2) + \frac{1}{6} \ln (5)$[/tex]
Therefore comparing both the sides,
[tex]$g(2) = \ln (2), g(5) = \ln(5), g(1) = 0 = \ln(1)$[/tex]
[tex]$g(X) = \ln(x)$[/tex]
Also, [tex]$g(1) =\ln(1)=0, g(2)= \ln(2) = 0.6931, g(5) = \ln(5) = 1.6094$[/tex]
b).
We known that [tex]$E[g(x)] = \sum g(x)p(x)$[/tex]
∴ [tex]$g(1).P(X=1) +g(2).P(X=2)+g(5).P(X=5) = \frac{1}{2}e^t+ \frac{2}{3}e^{2t}+ \frac{5}{6}e^{5t}$[/tex]
[tex]$g(1).\frac{1}{2} +g(2).\frac{1}{3}+g(5).\frac{1}{6 }= \frac{1}{2}e^t+ \frac{2}{3}e^{2t}+ \frac{5}{6}e^{5t}$[/tex]
Therefore on comparing, we get
[tex]$g(1)=e^t, g(2)=2e^{2t}, g(5)=5e^{5t}$[/tex]
∴ [tex]$g(X) = xe^{tx}$[/tex]
