Suppose X and N dependent random variables, where X is a discrete random variabl
ID: 3177497 • Letter: S
Question
Suppose X and N dependent random variables, where X is a discrete random variable are two in that is equally likely to be -1 or 1 (and cannot be any other value) and N has the Laplacian distribution where f_N(y) = e^-2|y|. Suppose Z = X + N. (a) What is the joint probability distribution function of X and N, F_XN(x, y? (b) Find the conditional density function of x given N = y, f_x (x|N = y) for all y element R. (c) Find the conditional density function of N given X = x, f_N (y|X = x) for all x elenent R. (d) What is the condition density function of Z given N = y, f_z(z|N = y) for all y elenent R? Using f_z(z|N = y), show the density function of Z is f_z(z) = 1/2{e^-2|x-1| + e^-2|x+1|}. (e) Find P(X = 1|Z = 1/3).Explanation / Answer
1.
Joint distribution function is f(x,y) = Xexp-2y : y is always positive .Taking –ve value will cause probability to be –ve which is not practical
N total
X 1 2 3
1 0.14 0.02 0.025 0.185
Total 0.14 0.02 0.025
2.
Conditional distribution of X when N = y
F(X) = fx/fn ; take value in such a manner that prob < 1
N
X 1 2 3
1 0.14 0.02 0.025
3.
Conditional distribution of N when X =1 ; -1 avoided because it will cause probability to be negative .
Fn = fn / fx ;
N
X 1 2 3
1 0.14 0.02 0.025
4. .
Z = X + N = 1 + exp-2y ; y always positive due to mod operator
Z practically occurs at high values of y typicall at > 1000 .
Conditional density function z ; fz(X,N)/f(N)
= (1 + exp-2y)/exp-2y
= 1 + exp-2y .
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