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). 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)

Answer 1

Answer:

a. $g(x) = ln(x)$.

b. $g(x) = x*e^{xt}$

Step-by-step explanation:

By the definition, the expected value of a random variable X with probability mass function p is given by $\sum P(X=x)\cdot x$ where the sum runs over all the posible values of X. Given a function g, the random variable Y=g(X) is defined. Note that the function g induces a probability mass function P' given by P'(Y=k) = P(X=g^{-1}(k)) when the function g is bijective.

a. Note that for 1/3ln(2)+1/6ln(5) by choosing the function g(x) = ln(x) the expression coincides with E(g(x)), because if Y = g(x)  then E(Y) = P'(Y=1)*ln(1)+P'(Y=2)*ln(2)+P'(Y=5)*ln(5) = P(X=1)*ln(1)+P(X=2)*ln(2)+P(X=5)*ln(5).

b. On the same fashion, the function g(x) = xe^{xt} fullfills the expression of E[g(X)]