Steven G
Elite Member
- Joined
- Dec 30, 2014
- Messages
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I can not see where I am going wrong. Hopefully it is just a simple arithmetic problem. Please let me know where i went wrong.
I have 2 random variables which are mutually independent.
The pmf for R and T are PR(r) = (.1)(.9)r-1 and PT(t) = (.2)(.8)t-1
So PRT(r,t)=(.02)(.9)r-1(.8)t-1=(1/36)(.9)r(.8)t
I want to find P(R < T).
\(\displaystyle \dfrac{1}{36}\begin{align*} \sum\limits_{t = 1}^\infty {\left[ {\sum\limits_{r = 0}^{t-1} {\left( {(.9)^r(.8)^t} \right)} } \right]}\end{align*}\)=\(\displaystyle \dfrac{1}{36}\begin{align*} \sum\limits_{t = 1}^\infty {(.8)^t}{\left[ {\sum\limits_{r = 0}^{t-1} {\left( {(.9)^r} \right)} } \right]}\end{align*}\) = \(\displaystyle \dfrac{1}{36}\begin{align*} \sum\limits_{t = 1}^\infty {(.8)^t}{[\dfrac {1-.9^t}{.1}]}\end{align*}\)=\(\displaystyle \dfrac{5}{18}\begin{align*} \sum\limits_{t = 1}^\infty {[(.8)^t-(.72)^t]}\end{align*}\)= \(\displaystyle \dfrac{5}{18}(\dfrac{.8}{1-.8} - \dfrac{.72}{1-.72}) = \dfrac{5}{18}(4 - \dfrac{72}{28}) = \dfrac{25}{63}\)
I have 2 random variables which are mutually independent.
The pmf for R and T are PR(r) = (.1)(.9)r-1 and PT(t) = (.2)(.8)t-1
So PRT(r,t)=(.02)(.9)r-1(.8)t-1=(1/36)(.9)r(.8)t
I want to find P(R < T).
\(\displaystyle \dfrac{1}{36}\begin{align*} \sum\limits_{t = 1}^\infty {\left[ {\sum\limits_{r = 0}^{t-1} {\left( {(.9)^r(.8)^t} \right)} } \right]}\end{align*}\)=\(\displaystyle \dfrac{1}{36}\begin{align*} \sum\limits_{t = 1}^\infty {(.8)^t}{\left[ {\sum\limits_{r = 0}^{t-1} {\left( {(.9)^r} \right)} } \right]}\end{align*}\) = \(\displaystyle \dfrac{1}{36}\begin{align*} \sum\limits_{t = 1}^\infty {(.8)^t}{[\dfrac {1-.9^t}{.1}]}\end{align*}\)=\(\displaystyle \dfrac{5}{18}\begin{align*} \sum\limits_{t = 1}^\infty {[(.8)^t-(.72)^t]}\end{align*}\)= \(\displaystyle \dfrac{5}{18}(\dfrac{.8}{1-.8} - \dfrac{.72}{1-.72}) = \dfrac{5}{18}(4 - \dfrac{72}{28}) = \dfrac{25}{63}\)
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