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STAT 333 a1sol

# STAT 333 a1sol - Stat 333 Assignment 1 Solutions Fall 2011...

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Stat 333 - Assignment 1 Solutions - Fall 2011 1. (a) We will define random variables I j to indicate the state of computer j , j = 1 , 2 , 3 , 4 . Therefore, I j = 1 , if computer j is isolated 0 , if computer j is not isolated Computer j is isolated iff all of the connections to this computer are broken. There are 3 connections to computer j . Therefore, the probability that computer j is iso- lated is P ( I j = 1) = p 3 . (b) The probability that computers 1 and j ( j = 2 , 3 , 4 ) are isolated is the probability that all connections to these computers are broken. There are 5 connections to computers 1 and 2. Therefore, the probability that computers 1 and 2 is isolated is P ( I 1 = 1 , I 2 = 1) = p 5 . There are 4 connections to computers 1 and 3. Therefore, the probability that computers 1 and 3 is isolated is P ( I 1 = 1 , I 3 = 1) = p 4 . There are also 6 connections to computers 1 and 4. Therefore, the probability that computers 1 and 4 is isolated is P ( I 1 = 1 , I 4 = 1) = p 6 . (c) Let T denote the total number of isolated computers. Then T = I 1 + I 2 + I 3 + I 4 . We know that, for j = 1 , 2 , 3 , 4 , E ( I j ) = P ( I j = 1) = p 3 and V ar ( I j ) = P ( I j = 1) · [1 - P ( I j = 1)] = p 3 - p 6 . Additionally, for j = k , Cov ( I j , I k ) = E ( I j · I k ) - E ( I j ) E ( I k ) = E ( I j · I k ) - p 3 · p 3 = E ( I j · I k ) - p 6 We can find that for j = k , E ( I j I k ) = P ( I j = 1 , I k = 1) and this depends on the values of j and k . E ( I 1 I 3 ) = E ( I 2 I 4 ) = p 4 = Cov ( I 1 , I 3 ) = Cov ( I 2 , I 4 ) = p 4 - p 6 . E ( I 1 I 2 ) = E ( I 3 I 4 ) = p 5 = Cov ( I 1 , I 2 ) = Cov ( I 3 , I 4 ) = p 5 - p 6 . E ( I 1 I 4 ) = E ( I 2 I 3 ) = p 6 = Cov ( I 1 , I 4 ) = Cov ( I 2 , I 3 ) = p 6 - p 6 = 0 . Therefore, E ( T ) = E ( I 1 ) + E ( I 2 ) + E ( I 3 ) + E ( I 4 ) = 4 · p 3 and V ar ( T ) = V ar ( I 1 + I 2 + I 3 + I 4 ) = { V ar ( I 1 ) + V ar ( I 2 ) + V ar ( I 3 ) + V ar ( I 4 ) } + 2 j<k Cov ( I j , I k ) = 4( p 3 - p 6 ) + 2 2( p 4 - p 6 ) + 2( p 5 - p 6 ) + 2(0) = 4( p 3 + p 4 + p 5 - 3 p 6 ) . 1

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(d) From part c), we see that the mean number of computers decrease when the p de- creases. This makes sense since we would expect fewer computers to be isolated if the chance of a broken connection is smaller. Notice that the variance is zero if p = 0 or p = 1 . For p < 0 . 808061 , an increase in p results in an increase in the variance whereas for p > 0 . 808061 , an increase in p leads to a decrease in the variance. To explain this behaviour intuitively, think for example that we have 10 rooms each with this computer network. If p is very large, in each room, we will constantly expect to have a large number of isolated computers (for example, 3 to 4 in each room). If p is very small, in each room, we will constantly expect to have a small number of isolated computers (for example, 0 to 1 in each room). Therefore, the variability is small for each of these cases. If
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