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 :

$P(X=1)=\frac{1}{2}, P(X=2)=\frac{1}{3}, P(X=5)=\frac{1}{6}$

a). We know :

    $E[g(x)] = \sum g(x)p(x)$

So,  $g(1).P(X=1) + g(2).P(X=2)+g(5).P(X=5) = \frac{1}{3} \ln (2) + \frac{1}{6} \ln(5)$

       $g(1).\frac{1}{2} + g(2).\frac{1}{3}+g(5).\frac{1}{6} = \frac{1}{3} \ln (2) + \frac{1}{6} \ln (5)$

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

∴ $g(1).P(X=1) +g(2).P(X=2)+g(5).P(X=5) = \frac{1}{2}e^t+ \frac{2}{3}e^{2t}+ \frac{5}{6}e^{5t}$

   $g(1).\frac{1}{2} +g(2).\frac{1}{3}+g(5).\frac{1}{6 }= \frac{1}{2}e^t+ \frac{2}{3}e^{2t}+ \frac{5}{6}e^{5t}$

Therefore on comparing, we get

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

∴ $g(X) = xe^{tx}$