Question

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O Let X and Y be independent random variables with a discrete uniform distribution, i.e., with probability mass functions for k = 1, px(k) = py (k) =-, N. Use the addition rule for discrete random variables on page 152 to determine the probability mass function of Z -X+Y for the following two cases. a. Suppose N = 6, so that X and Y represent two throws with a die. Show that 2,... ,6, for k 36 13-k P2(k) = P(X + Y = k) = for k-7,., 12. 36 You may check this with Quick exercise 11.1 b. Determine the expression for pz(k) for general N.

Answer 1

a.

N=6 So value of Z = k , k=2,3,....,12

X and Y are independent.

p_z(2) = P(X+Y=2) = P(X=1,Y=1) = P(X=1)P(Y=1) = (1/6)(1/6) = 1/36 = (2-1)/36

p_z(3) = P(X+Y=3) = P(X=1,Y=2) + P(X=2,Y=1) = P(X=1)P(Y=2) + P(X=2)P(Y=1) = (1/6)(1/6) + (1/6)(1/6) = 2/36 = (3-1)/36

p_z(4) = P(X+Y=4) = P(X=1,Y=3) + P(X=2,Y=2) + P(X=3,Y=1) = P(X=1)P(Y=3) + P(X=2)P(Y=2) + P(X=3)P(Y=1) = (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) = 3/36 = (4-1)/36

p_z(5) = P(X+Y=5) = P(X=1,Y=4) + P(X=2,Y=3) + P(X=3,Y=2) + P(X=4,Y=1) = P(X=1)P(Y=4) + P(X=2)P(Y=3) + P(X=3)P(Y=2) + P(X=4)P(Y=1) = (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) = 4/36 = (5-1)/36

p_z(6) = P(X+Y=6) = P(X=1,Y=5) + P(X=2,Y=4) + P(X=3,Y=3) + P(X=4,Y=2) + P(X=5,Y=1) = P(X=1)P(Y=4) + P(X=2)P(Y=4) + P(X=3)P(Y=3) + P(X=4)P(Y=2) + P(X=5)P(Y=1) = (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) = 5/36 = (6-1)/36

p_z(7) = P(X+Y=7) = P(X=1,Y=6) + P(X=2,Y=5) + P(X=3,Y=4) + P(X=4,Y=3) + P(X=5,Y=2) + P(X=6,Y=1) = P(X=1)P(Y=6) + P(X=2)P(Y=5) + P(X=3)P(Y=4) + P(X=4)P(Y=3) + P(X=5)P(Y=2) + P(X=6)P(Y=1) = (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) = 6/36 = (13-7)/36

p_z(8) = P(X+Y=8) = P(X=2,Y=6) + P(X=3,Y=5) + P(X=4,Y=4) + P(X=5,Y=3) + P(X=6,Y=2) = P(X=2)P(Y=6) + P(X=3)P(Y=5) + P(X=4)P(Y=4) + P(X=5)P(Y=3) + P(X=6)P(Y=2) = (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) = 5/36 = (13-8)/36

p_z(9) = P(X+Y=9) = P(X=3,Y=6) + P(X=4,Y=5) + P(X=5,Y=4) + P(X=6,Y=3) = P(X=3)P(Y=6) + P(X=4)P(Y=5) + P(X=5)P(Y=4) + P(X=6)P(Y=3) = (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) = 4/36 = (13-9)/36

p_z(10) = P(X+Y=10) = P(X=4,Y=6) + P(X=5,Y=5) + P(X=6,Y=4) = P(X=4)P(Y=6) + P(X=5)P(Y=5) + P(X=6)P(Y=4) = (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) = 3/36 = (13-10)/36

p_z(11) = P(X+Y=11) = P(X=5,Y=6) + P(X=6,Y=5) = P(X=5)P(Y=6) + P(X=6)P(Y=5) = (1/6)(1/6) + (1/6)(1/6) = 2/36 = (13-11)/36

p_z(12) = P(X+Y=12) = P(X=6,Y=6) = P(X=6)P(Y=6) = (1/6)(1/6) = 1/36 = (13-12)/36

$\texttt{Thus it can be seen that :- }$

$p_z(k)=\texttt P(X+Y=k)=\left\{\begin{matrix} \frac{k-1}{36} & for\;k=2,......,6\\ \frac{13-k}{36} & for\;k=7,......,12 \end{matrix}\right.$

.

b.

For general N ; value of Z = k , k = 2,3,....,2N

$\texttt{For k = 2,3,....,N }\;\;\;\;:-$

$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! p_z(k)=\texttt P(X+Y=k)=\sum_{i=1}^{k-1}\texttt P(X=i,Y=k-i)=\sum_{i=1}^{k-1}\texttt P(X=i)\texttt P(Y=k-i)=\sum_{i=1}^{k-1}\left ( \frac{1}{\texttt N} \right )\left ( \frac{1}{\texttt N} \right )=\frac{k-1}{\texttt N^2}$

$\texttt{For k = N+1,N+2,....,2N }\;\;\;\;:-$

$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! p_z(k)=\texttt P(X+Y=k)=\sum_{j=k-\texttt N}^{\texttt N}\texttt P(X=j,Y=k-j)=\sum_{j=k-\texttt N}^{\texttt N}\texttt P(X=j) \texttt P(Y=k-j)=\sum_{j=k-\texttt N}^{\texttt N}\left ( \frac{1}{\texttt N} \right )\left ( \frac{1}{\texttt N} \right )=\frac{2\texttt N-k+1}{\texttt N^2}$