prelim 2 solutions

# B we will show that x y is binomial 2 n 1 2 random

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(b) We will show that X + Y is Binomial (2 n, 1 2 ) random variable, i.e., P ( X + Y = k )= braceleftbigg 1 4 n ( 2 n k ) , if 0 k 2 n 0 , otherwise This result can be established by noticing that a Binomial ( n,p ) random variable can be viewed as the sum of n independent Bernoulli ( p ) random variables. Hence the sum of two independent Binomial ( n,p ) random variables can be regarded as the sum of 2 n independent Bernoulli ( p ) random variables. This shows that the sum of two independent Binomial ( n,p ) random variables is a Binomial (2 n,p ) random variable. We can arrive to this result also through calculations. Indeed, the random variables X and Y are Binomial ( n, 1 2 ) . This means that, P ( X = i )= P ( Y = i )= braceleftbigg ( n i ) 1 2 i 1 2 n - i = ( n i ) 1 2 n , if 0 i n 0 , otherwise To compute the PMF of X + Y , we note first that P ( X + Y = k )= k summationdisplay i =0 P ( X = i,Y = k i ) = k summationdisplay i =0 P ( X = i ) P ( Y = k i ) , where in the second equality we used the independence between X and Y . Now if 0 k n , we have P ( X + Y = k )= k summationdisplay i =0 P ( X = i ) P ( Y = k i ) = 1 4 n k summationdisplay i =0 parenleftbigg n i parenrightbiggparenleftbigg n k i parenrightbigg . On the other hand, if n < k 2 n , we have P ( X = i ) = 0 for all i > n and P ( Y = k i ) = 0 for i < k n , this means P ( X + Y = k )= n summationdisplay i = k n P ( X = i ) P ( Y = k i ) = 1 4 n n summationdisplay i = k n parenleftbigg n i parenrightbiggparenleftbigg n k i parenrightbigg . 2

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Finally, since the maximum value for X and Y is n , we have and P ( X + Y = k )=0 for k > 2 n . In conclusion, P ( X + Y = k )= 1 4 n k i =0 ( n i )( n k i ) , if 0 k n 1 4 n n i = k n ( n i )( n k i ) , if n +1 k 2 n 0 , otherwise This can be written in a more compact from as P ( X + Y = k )= braceleftBigg 1 4 n min( k,n ) i =max(0 ,k n ) ( n i )( n k i ) , if 0 k 2 n 0 , otherwise The rest of this problem can be solved using the formula above. In the following, we will simplify this result using combinatorial identities (note: you do not need to know these formulas in the exam). In combinatorics, Vandermonde’s convolution states the following parenleftbigg 2 n k parenrightbigg = k summationdisplay i =0 parenleftbigg n i parenrightbiggparenleftbigg n k i parenrightbigg . (1) Consequently, for 0 k n , we have P ( X + Y = k )= 1 4 n parenleftbigg 2 n k parenrightbigg . Now for n +1 k 2 n , making the change of variable j = i + n k , we have n summationdisplay i = k n parenleftbigg n i parenrightbiggparenleftbigg n k i parenrightbigg = 2 n k summationdisplay j =0 parenleftbigg n j + k n parenrightbiggparenleftbigg n n j parenrightbigg , = 2 n k summationdisplay j =0 parenleftbigg n j parenrightbiggparenleftbigg n 2 n k j parenrightbigg . where in the second equality we have used the facts that ( n j + k n ) = ( n 2 n k j ) and ( n n j ) = ( n j ) . Now using Vandermonde’s identity, this becomes n summationdisplay i = k n parenleftbigg n i parenrightbiggparenleftbigg n k i parenrightbigg = parenleftbigg 2 n 2 n k parenrightbigg = parenleftbigg 2 n k parenrightbigg .
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