Geometric Random Variable.

Oct 2009
31
2
Hey all i was stuck on this geometric random variable question.



Part (i)

(a) pmf = \(\displaystyle p(k) = (1-p)^{k-1} \cdot p \)
(b) is a simple proof.


Part (ii)

(a) \(\displaystyle p(X \ge 4) = (1 -\dfrac{1}{3})^{4-1} = (\dfrac{2}{3})^3 = \dfrac {8}{27} \)

(b) & (c) i dont have an idea of how to attempt, any help is greatly appricated.

Thanks guys
 

matheagle

MHF Hall of Honor
Feb 2009
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2b or not 2b, is that the question, just write out a few cases...

\(\displaystyle P(X=2)=P(RR)=(1/3)^2\)

\(\displaystyle P(X=3)=P(BRR)+P(RBR)=2(1/3)^2(2/3)\)

\(\displaystyle P(X=4)=P(BBRR)+P(BRBR)+P(RBBR)=3(1/3)^2(2/3)^2\)

It looks like \(\displaystyle P(X=k)=(k-1)(1/3)^2(2/3)^{k-2}\), \(\displaystyle k=2,3,4....\)
 
Last edited:

matheagle

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Feb 2009
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This is a valid distribution...

\(\displaystyle p^2\sum_{k=2}^{\infty}(k-1)(1-p)^{k-2}\)

\(\displaystyle =-p^2{d\over dp}\left(\sum_{k=2}^{\infty}(1-p)^{k-1}\right)\)

\(\displaystyle =-p^2{d\over dp}\left(\sum_{j=0}^{\infty}(1-p)^{j+1}\right)\)

\(\displaystyle =-p^2{d\over dp}\left({1-p\over 1-(1-p))}\right)\)

\(\displaystyle =-p^2{d\over dp}\left(p^{-1}-1\right)\)

\(\displaystyle =-p^2\left(-p^{-2}\right)\)

\(\displaystyle =1\)
 

matheagle

MHF Hall of Honor
Feb 2009
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I was waiting for someone to do 2c...
This is a finite rv, with W=1,2,...,11.

\(\displaystyle P(W=1)=P(R)= {5\over 15}={1\over 3}\)

\(\displaystyle P(W=2)=P(BR)= \left({10\over 15}\right)\left({5\over 14}\right) \)

\(\displaystyle P(W=3)=P(BBR)= \left({10\over 15}\right) \left({9\over 14}\right)\left({5\over 13}\right) \)