Q of Geometric Distribution

2012-04-04 5:01 pm
The random variable X has geometric distribution with pmf p(x)= q^x p, x=1,2,..., 0<p<1 where q=1-p

(note that captioned pmf is amended as question)

1. Explain why P(X is odd) = qP(X is even) and hence or otherwise show that P(X is odd) = q / (1+ q)

2. The random variable Y is independent of X and has pmf p(y)= qp^ y, y=1,2,...

Write down P(X=k and Y=k) for an arbitrary non-negative integer k and hence show that

P(X=Y)= p(1-p)/ [1-p(1-p)]

Using calculus, find the value of p that maximizes this expression and hence deduce the maximum possible value of P(X=Y) is 1/3.

(Note that you are not required to show any turning point that you locate is a maximum)

(I look into for a long while but have no idea yet.....please kindly assist if anyone expert is familiar with...Thanks a lot!)


更新1:

it is very appreciated and thanks for your prompt answer, myisland8132. But I can't get the MAX∣x=0 for x=p(1-p) as 1/4 by equating the f(x)=0. Would you kindly elaborate? Thanks a lot!

更新2:

sorry the value i can't deduce is the largest value for x=p(1-p) by equating the f'(x)=0 as 1/4. Please ignore the previous description. Thanks!

回答 (1)

2012-04-04 7:32 pm
✔ 最佳答案
1 P(X is odd) = qp + q^3p + ...= p(q + q^3 + ...)= qp(1 + q^2 + ....)= qP(X is even)As P(X is odd) + P(X is even) = 1P(X is odd) + (1/q)P(X is odd) = 1P(X is odd) = q/(1 + q)2 P(X=k and Y=k) = P(X = k)P(Y = k)= q^kp * qp^k= q^(k + 1)p^(k + 1)= [p(1 - p)]^(k + 1)P(X = Y)= ∑ [p(1 - p)]^(k + 1)= p(1 - p)/[1 - p(1 - p)]Let f(x) = x/(1 - x)f'(x) = 1/(1 - x)^2 > 0 => f(x) is increasingBut the largest value of p(1 - p) is 1/4So, the maximum value of P(X = Y) = (1/4)/(1 - 1/4) = 1/3


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