Question

(e) Suppose that 4 balls are placed sequentially into one of 5 bins, where the bin for each ball is selected at random. For i = 1, 2,3,4,5, define the indicator variables 1, if the ith bin is empty; 0, otherwise Xi _ Then the number of empty boxes is given by X = X\+X2+ X3 + X4+X5, and we learned from week #7 lecture notes and midterm II that Xi ~ Bernoulli(p) with p = (1 - 2)4 = 0.4096 (i) For Bernoulli random variable X4, find the mean E(X;) and variance Var(X;) (ii Find the mean E(X), the expected number of empty bins. (iii For 1i /j< 5, determine the covariance of X; and X; Hints: E(XiX;) = P(X4 = 1, X; = 1) = 0.64 = 0.1296. Why? 54 (iv) Use (d) to find the variance of X = Xı + X2 + X3 + X4 + X5, the number of empty bins.

Answer 1

(i)

We are given that:

X_i ~ Bernoulli(p) with p = $\small \left ( 1-\frac{1}{5} \right )^4$ = 0.4096

Now, the mean of X_i is given by:

$\small \begin{align*} E(X_i) &= \sum_{x=0}^{1}x *P(X_i=x) \\ &= 0*P(X_i =0 ) + 1*P(X_i =1 ) \\ &= 0 + p \\ &= p \\ &= \textbf{0.4096 \ \ \ \ [Answer]} \end{align*}$

Now, before we calculate the variance consider:

$\small \begin{align*} E(X_i^2) &= \sum_{x=0}^{1}x^2 *P(X_i=x) \\ &= 0^2*P(X_i =0 ) + 1^2*P(X_i =1 ) \\ &= 0 + p \\ &= p \end{align*}$

Thus, variance is given by:

$\small \begin{align*} Var(X_i) &= E(X_i^2) - E^2(X_i) \\ &= p - p^2 \\ &= p(1-p) \\ &= 0.4096*(1-0.4096) \\ &= \textbf{0.241828 \ \ \ \ \ [Answer]} \end{align*}$

(ii)

$\small \begin{align*} E(X) &= E(X_1+X_2+X_3+X_4+X_5) \\ &= E(X_1) + E(X_2) + E(X_3) +E(X_4)+E(X_5) \\ &= 5*E(X_i) \\ &= 5*0.4096 \\ &= \textbf{2.048 \ \ \ \ [Answer]} \end{align*}$

(iii)

Before we calculate the covariance, consider (for i $\small \ne$ j):

$\small \begin{align*} E(X_iX_j) &= 0*P(X_iX_j=0) + 1*P(X_iX_j=1) \\ &= P(X_iX_j=1) \\ &= P(X_i =1,X_j=1) \ \ \ \ \text{[Because }X_iX_j=1 \text{ iff }X_i=1 \text{ and }X_j=1] \\ &= P(\text{both ith and jth bins are empty}) \\ &= P(\text{1st ball is placed in any of the three bins excpet ith and jth bin})*... \\ & \ \ \ \ *P(\text{4th ball is placed in any of the three bins excpet ith and jth bin}) \\ &= \frac{3}{5} *\frac{3}{5}*\frac{3}{5}*\frac{3}{5} \\ &= \left ( \frac{3}{5} \right )^4 \\ &= 0.1296 \end{align*}$

$\small \begin{align*} Cov(X_i,X_j) &= E(X_iX_j) - E(X_i)E(X_j) \\ &= 0.1296 - 0.4096*0.4096 \\ &= \textbf{-0.038172 \ \ \ \ \ \ \ [Answer] } \end{align*}$

(iv)

$\small \begin{align*} Var(X) &= Var(X_1+X_2+X_3+X_4+X_5) \\ &= [Var(X_1) + ...+Var(X_5)] + 2*[Cov(X_1,X_2) + Cov(X_1,X_3) + ... + Cov(X_4,X-5)] \\ &= 5*Var(X_1) + 2*[10*Cov(X_i,X_j)] \\ &= 5*0.241828 + 20*(-0.038172) \\ &= \textbf{0.445696 \ \ \ \ \ \ [Answer]} \end{align*}$

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