0

A discrete random variable $X$ takes value $0,1,2, \ldots n$ with frequency $\binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n}$. Find the variance.

I have calculated the mean as such $$\mathbb{P}(X)= \binom{n}{x}/2^n$$ Therefore, $$\begin{split} \mathbb{E}(X) &= \binom{n}{1}/2^n+ \binom{n}{2}/2^n + \ldots + \binom{n}{n}/2^n\\ &= \frac{\binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n}}{2^n}\\ &= \frac{n \cdot 2^{n-1}}{2^n}\\ &= n/2 \end{split} $$ But $$\mathbb{E}(X^2) = \binom{1}{1}/2^n+ \binom{4}{2}/2^n + \ldots $$

I am not able to get an algebraic expression for this series. I know the answer is $n/4$.

gt6989b
  • 54,422
kangkan
  • 219
  • I tried to make this readable, please verify this is what you intended. – gt6989b Feb 27 '14 at 19:44
  • 1
    If this is the binomial with probability of success $1/2$, the probability that $X=k$ is, in your notation, ${}_n\text{C}_k$. The expression for the mean is as of now not correct, though $n/2$ is correct. The expression for $E(X^2)$ is not right. – André Nicolas Feb 27 '14 at 19:45

1 Answers1

1

So the mean should be $$ \mathbb{E}[X] = \sum_{k=1}^n \binom{n}{k} k /2^n = \frac{1}{2^n} \sum_{k=1}^n \binom{n}{k} k $$ and similarly $$ \mathbb{E}[X^2] = \frac{1}{2^n} \sum_{k=1}^n \binom{n}{k} k^2 $$

Can you take it from here?

HINT one more word about how to handle such sums. You derived the probability distribution correctly, so you know that $\sum_{k=0}^n \binom{n}{k} = 2^n$. A more general form of that is the Binomial Theorem $$ (1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k $$ (you get your result for $x=1$). Note that taking the first derivative we get $$ n(1+x)^{n-1} = \sum_{k=0}^n \binom{n}{k}kx^{k-1} $$ and substitution of $x=1$ makes things much more simple. Try another differentiation to handle the second moment.

gt6989b
  • 54,422