Expected Value of Binomial Distribution

Question

What is the expected value of the absolute value of the difference between the number of incoming tails and the number of incoming heads when a coin is tossed 5 times?

Here is what I think : Let $X$ is random variable of number of incoming heads then $p=\frac{1}{2}$ and $n=5$ so $X\sim Binom(5,\frac{1}{2})$

$X$ and $Y=5-X$ ($Y$ is random variable of number of incoming tails) then $E(K)=E(\left|X-(5-X)\right|)=E(\left|2X-5\right|)=E(2X-5)=2E(X)-5=2.\frac{5}{2}-5=0$ where $K=\left|X-Y\right|=\left|2X-5\right|$ and $E(X)=n.p=5.\frac{1}{2}=\frac{5}{2}$

But the right answer is $\frac{15}{8}$. Where am I making a mistake? Any help will be appreciated.

Answer 1

Your random variable $Z=|X-Y|$ can take only the values

$$\{1,3,5\}$$

with probabilities

$$\left\{\frac{20}{32};\frac{10}{32};\frac{2}{32}\right\}$$

This because

$Z=5$ when you get 5 consecutive tails (or heads)

$Z=3$ when you get 4 tails (or heads)

$Z=1$ when you get 2 tails (or heads)

No other situations are possible.

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Thus its expectation is

$$\mathbb{E}[Z]=1\times \frac{20}{32}+3\times\frac{10}{32}+5\times\frac{2}{32}=\frac{15}{8}$$

Answer 2

Since $X\sim Binom(5,\frac{1}{2})$, the range of $X$ is $\{0,1,2,3,4,5\}$, which implies that $2X - 5$ is not necessarily always positive. Therefore in your proof, it is not true that $E(\left|2X - 5\right|) = E(2X - 5)$.