Question

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). Your answer should give at least the values g(k) for all possible values k of X, but you can also specify g on a larger set if possible. (b) Let t be some real numb

Answer 1

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 :

`$P(X=1)=(1)/(2), P(X=2)=(1)/(3), P(X=5)=(1)/(6)$`

a). We know :

`$E[g(x)] = \sum g(x)p(x)$`

So,
`$g(1).P(X=1) + g(2).P(X=2)+g(5).P(X=5) = (1)/(3) \ln (2) + (1)/(6) \ln(5)$`

`$g(1).(1)/(2) + g(2).(1)/(3)+g(5).(1)/(6) = (1)/(3) \ln (2) + (1)/(6) \ln (5)$`

Therefore comparing both the sides,

`$g(2) = \ln (2), g(5) = \ln(5), g(1) = 0 = \ln(1)$`

`$g(X) = \ln(x)$`

Also,
`$g(1) =\ln(1)=0, g(2)= \ln(2) = 0.6931, g(5) = \ln(5) = 1.6094$`

b).

We known that
`$E[g(x)] = \sum g(x)p(x)$`

∴
`$g(1).P(X=1) +g(2).P(X=2)+g(5).P(X=5) = (1)/(2)e^t+ (2)/(3)e^(2t)+ (5)/(6)e^(5t)$`

`$g(1).(1)/(2) +g(2).(1)/(3)+g(5).(1)/(6 )= (1)/(2)e^t+ (2)/(3)e^(2t)+ (5)/(6)e^(5t)$$`

Therefore on comparing, we get

`$g(1)=e^t, g(2)=2e^(2t), g(5)=5e^(5t)$`

∴
`$g(X) = xe^(tx)$`