Question

8. For each i = 1, 2, ..., 10, Xi is a random variable that gives 0 or 1 if the ith toss of a fair coin came up T or H, respectively. Let X = X1 + X2 + ... + X10 a. Compute the expected value E(X) and variance V(X) of X. [5] b. What is the probability function of X? (10)

Answer 1

a.

Note that X_i's (i=1,2,...,10) are independent and identically distributed. They are independent because each toss of a coin is independent and they are identical because on each toss the probability of head/tail is same.

Moreover, note that on each toss of a fair coin P(H) = P(T) = 1/2

Now, we find the expected value of X_i's (i=1,2,...,10):

$\begin{align*} E(X_i) &= 0*P(X_i=0) + 1*P(X_i = 1) \\ &= 0*P(T) + 1*P(H) \\ &\text{[Since, }X_i = 0 \text{ when ith toss comes up T and }X_i = 1 \text{ when ith toss comes up H]} \\ &= 0*\frac{1}{2} + 1*\frac{1}{2} \\ &= 0 + \frac{1}{2} \\ &= \frac{1}{2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ; i=1,2,...,10 \end{align*}$

Now, we find:

$\begin{align*} E(X_i^2) &= 0^2*P(X_i=0) + 1^2*P(X_i = 1) \\ &= 0*P(T) + 1*P(H) \\ &\text{[Since, }X_i = 0 \text{ when ith toss comes up T and }X_i = 1 \text{ when ith toss comes up H]} \\ &= 0*\frac{1}{2} + 1*\frac{1}{2} \\ &= 0 + \frac{1}{2} \\ &= \frac{1}{2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ; i=1,2,...,10 \end{align*}$

Thus, the variance of X_i's (i=1,2,...,10) is given by:

$\begin{align*} Var(X_i) &= E(X_i^2) - \left[E(X_i) \right ]^2 \\ &= \frac{1}{2} - \left[\frac{1}{2} \right ]^2 \\ &= \frac{1}{2} - \frac{1}{4} \\ &= \frac{1}{4} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ; i=1,2,...,10 \end{align*}$

Now, we find the expected value and variance of X:

E(X) = E(X1 + X2 + ... + X 10) E(X1) + E(X2) + ... + E(X10) 1 = + 2 1 1 + + 10 times 1 = 10 2 = 5 [ANSWER V(X)= V(X1 + X2 + ... + X10) = V(X1) + V(X2) + ... +V(X10) [Since, Xs are independent] 1 1 + 1 + + 10 times 1 = 10* 4 2.5 [ANSWER]

b.

We observe that X_i (i=1,2,...,10) is equal to 1 if the ith toss comes up H, otherwise it is equal to 0.

Thus, is just the number of heads we get on 10 tosses of a fair coin. Now, since there is a fixed number of tosses (equal to 10), each toss has two outcomes (H or T) and on each toss we get H with probability 1/2 independent of other tosses, thus we can conclude that:
X ~ Binomial(n = 10, p = 1/2)

X = X1 + X2 + ... + X 10

Thus, the probability function of X is given by:
$\begin{align*} \bf P(X=x) &= \binom{n}{x}p^x (1-p)^{n-x} \ \ \ \ \ \ \ \ \ \ \ &&;x=0,1,2,...,n \\ &= \binom{10}{x}*\left(\frac{1}{2} \right )^x *\left(\frac{1}{2} \right )^{10-x} \ \ \ \ \ \ \ \ \ \ \ &&;x=0,1,2,...,10 \\ &= \boldsymbol{\binom{10}{x}* \left(\frac{1}{2} \right )^{10} \ \ \ \ \ \ \ \ \ \ \ }&&\boldsymbol{;x=0,1,2,...,10} \\ &\textbf{[ANSWER]} \end{align*}$

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