Variance of the difference of two random variables

Question

I have the following problem: Suppose an unbiased coin is tossed 10 times. Let D be the random variable that denotes the number of heads minus the number of tails. What is the variance of D?

My solution is: Let X be the number of heads and Y be the number of tails, so $X~Bin(10,\frac{1}{2})$ and $Y~Bin(10,\frac{1}{2})$ and $D=X-Y$

Now $Var(D)=Var(X)+Var(Y)-2COV(X,Y)$ but since X and Y are Independent we have that covariance will be 0 i.e, $COV(X,Y)=0$

so $Var(D)=Var(X)+Var(Y)=2np(1-p)=2\cdot10\cdot\frac{1}{2}(1-\frac{1}{2})=5$

Where am i wrong in this? Any help is appreciated!

Hint. Try defining $X_i=1$ if toss $i$ is a head and $-1$ if it's a tail. — Blitzer, Mar 28 '24 at 07:17
Given $n$ tosses, heads + tails = $n$, so the number of heads AND number of tails cannot be independent variables, given $n$. — wolfies, Mar 28 '24 at 07:17

geetha290krm · Answer 1 · 2024-03-28T08:40:37.123

4

Note that $X+Y=10$. So $D=X-Y=X-(10-X)=2X-10$. This gives $var (D)=4 var (X)$. So $Var(D)=4\cdot np(1-p)=4\cdot 10\cdot \frac{1}{2}(1-\frac{1}{2})=10$.

edited Mar 28 '24 at 08:40

answered Mar 28 '24 at 07:17

geetha290krm

36,632

thank you @geetha290krm – econhead Mar 28 '24 at 07:23

Variance of the difference of two random variables

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