Answer to Question #100404 in Statistics and Probability for Nimalka

Question

Two 10p coins are tossed. The random variable X represents the total value of each coin lands heads up.
(a)Find E(X) and Var(X).

The random variables S and T are defined as follows:
S = X-10 and T = (1/2)X-5
(b)Show that E(S) = E(T).
(c)Find Var(S) and Var (T).

(d)
Susan and Thomas play a game using two 10p coins. The coins are tossed and Susan records her score using the random variable S and Thomas uses the random variable T. After a large number of tosses they compare their scores.
Comment on any likely differences or similarities.

Answer 1

a) Find the distribution of $X$

$$P(X=0)={1 \over 2}\cdot{1 \over 2}={1 \over4}$$

$$P(X=1)={1 \over 2}\cdot{1 \over 2}+{1 \over 2}\cdot{1 \over 2}={1 \over2}$$

$$P(X=2)={1 \over 2}\cdot{1 \over 2}={1 \over4}$$

$$\begin{array}{cc} \ \ \ \ \ \ \ \ \ \ \ \ \ \ x: & 0 & 1 & 2 \\ P(X=x): & {1 \over4} & {1 \over2} & {1 \over4} \end{array}$$

$$E(X)=0\cdot{1 \over4}+10\cdot{1 \over2}+20\cdot{1 \over4}=10$$

Or use symmetry $E(X)=10.$

$$Var(X)=0^2 \cdot{1 \over4}+10^2\cdot{1 \over2}+20^2\cdot{1 \over4}-10^2=50$$

b)

$$E(S)=E(X-10)=E(X)-10=10-10=0$$

$$E(T)=E({1 \over2}X-5)=E({1 \over2}X)-5={1 \over2}\cdot10-5=0$$

c)

$$Var(S)=Var(X)=50$$

$$Var(T)=({1 \over2})^2Var(X)={1 \over4}\cdot50=12.5$$

d)

Both random variables have expected value 0, so we would expect both Susan and Thomas to have a score of approximately 0.

The random variable S which represents Susan’s score has higher variance, meaning we should expect it to vary more.