Expected Value of Number of Tails Minus Number of Heads

Question

I have the following problem where, given $X_n$ is a random variable that equals the number of tails minus the number of heads when n fair coins are flipped, what is the expected value of $X_n$?

I am having a difficulty getting started on this problem. Could someone offer a suggestion as to how this problem should be modeled?

score 6 · Accepted Answer · answered Sep 21 '13 at 05:24

6

We can start and finish by saying that by symmetry the expected number is $0$.

Or let $X$ be the number of tails. So $n-X$ is the number of heads. Thus tails minus heads is $2X-n$. This has mean $2E(X)-n$, which is $0$. This second argument will also give the answer for a coin that is biased, having probability $p$ of tails. Then the expected value of the difference is $2pn-n$.

answered Sep 21 '13 at 05:24

André Nicolas

507,029

Why does 2E(X)-n equal 0? – shachna Jul 27 '16 at 04:21
1

@shachna: The number $X$ of tails has binomial distribution $p=1/2$. So $E(X)=(1/2)n$. Equivalently, $2E(X)-n=0$. – André Nicolas Jul 27 '16 at 04:25

score 1 · Answer 2 · answered Sep 21 '13 at 05:27

1

The key is linearity of expectation, and I suggest finding a way to write $X_n$ as a sum of simple 'indicators' corresponding to individual coin flips. So you would be thinking, "if I see a heads, then value to add is __.... and tails ......"

answered Sep 21 '13 at 05:27

Evan

3,861

Expected Value of Number of Tails Minus Number of Heads

2 Answers2