Calculating MGF for a random variable with pmf $P(X=x)=k\cdot( ^nC_{x})$

Question

The pmf of a random variable X is given by $P(X=x)=k\cdot( ^nC_{x})$, $x=0,1,2,...,n$, where k is a constant. The moment generating function $M_X(t)$ is

(A)$\dfrac{(1+e^t)^n}{2^n}$

(B)$\dfrac{2^n}{(1+e^t)^n}$

(C) $\dfrac{1}{2^n(1+e^t)^n}$

(D) $2^n(1+e^t)^n$

SOLUTION:

For a discrete probability mass function, $M_X(t)=\displaystyle\sum_{i=1}^\infty e^{tx_i}\, p_i$

$M_X(t)=\displaystyle\sum_{i=1}^\infty e^{tx_i}\,k\cdot( ^nC_{x})$

This looks like $M_X(t)=(1+e^t)^n$. How to deal with k here ?

The answer choices seem to suggest that $k=1/2$, but I am not sure.

Could anyone help me on this ?

Answer 1

You need to choose $k$ so that $$ k\binom{n}0+k\binom{n}1+\dots+k\binom{n}n=1 $$ Since $\binom{n}0+\dots+\binom{n}n=2^n$ (these both count the number of subsets of an $n$ element set), this means $k=\frac1{2^n}$.

Since $$ M_X(T)=\sum (e^t)^{x}\cdot k\cdot \binom{n}x=k\sum (e^t)^{x}\cdot \binom{n}x=k(1+e^t)^n $$ the answer is $\frac{(1+e^t)^n}{2^n}$.