Geometric Random Variable probability problem

AI Thread Summary
The discussion revolves around solving problems related to a geometric random variable X with a probability p = 0.1. The user successfully calculates the cumulative distribution function F_X(5) as approximately 0.40951 and the probability Pr(5 < X ≤ 11) as about 0.27668. They express uncertainty regarding the conditional probability Pr(X=7|5<X≤11) but provide a calculation leading to approximately 0.19208. The expected value E(X|3<X≤11) is computed as approximately 6.95, while E(X^2|3<X≤11) is found to be around 53.42, leading to a variance of approximately 5.07. The user seeks confirmation of their calculations and inquires about different methods for determining expected values.
probhelp150

Homework Statement


X is a geometric random variable with p = 0.1. Find:
##a. F_X(5)##
##b. Pr(5 < X \leq 11)##
##c. Pr(X=7|5<X\leq11)##
##d. E(X|3<X\leq11)##
##e. E(X^2|3<X\leq11)##
##f. Var(X|3<X\leq11)##

Homework Equations

The Attempt at a Solution


Can someone check my work and help me?
a.
##F_X(5) = P(X\leq5)=1-(1-p)^x##
##=1-(0.9)^5##
##=0.40951##b.
##Pr(5<X\leq11)=F_X(11)-F_X(5)##
##=0.2766794039##c. Not too sure about this one.
##Pr(X=7) = F_X(7) - F_X(6) = 0.0531441##
##Pr(5<X\leq11)=0.2766794039##
##\frac{Pr(X=7)}{Pr(5<X<\leq11)} = \frac{0.0531441}{0.2766794039}=0.192078##

d.
##Pr(3<X\leq11) = F_X(11)-F_X(3) = 0.41518940391##
##E(X|3<X\leq11) = \sum_{x=4}^{11} \frac{x*0.1*0.9^{x-1}}{Pr(3<X\leq11)}##
##=\sum_{x=4}^{11} \frac{x*0.1*0.9^{x-1}}{0.41518940391)}##
##=\frac{0.1}{0.41518940391}\sum_{x=4}^{11} {x*0.9^{x-1}}##
##=6.9533987498770702912469710479707##

e. Don't know how to do this one.
##Pr(3<X\leq11) = F_X(11)-F_X(3) = 0.41518940391##
##E(X^2|3<X\leq11) = \sum_{x=4}^{11} \frac{x^2*0.1*0.9^{x-1}}{Pr(3<X\leq11)}##
##=\sum_{x=4}^{11} \frac{x^2*0.1*0.9^{x-1}}{0.41518940391)}##
##=\frac{0.1}{0.41518940391}\sum_{x=4}^{11} {x^2*0.9^{x-1}}##
##=53.415557495820389902397015631005##
f. Variance = ##E[X^2] - E[X]^2 = 5.0658033210283859687442550652196##
 
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How is an expectation value defined?
 
Orodruin said:
How is an expectation value defined?
I believe expected value is the average. I reattempted the problem. Could you help me double check?
 
I get the same thing. Note that you would typically not need to quote so many decimals, it is essentially just distracting.
 
Orodruin said:
I get the same thing. Note that you would typically not need to quote so many decimals, it is essentially just distracting.
Are my answers for a,b,c right as well? I'm only using the "full" answer because I'm doing the work on my computer (Windows Calculator) and it was easier to copy/paste the answer. I know that it should be truncated on an exam or homework.

Is there another method of obtaining E[X] and E[X^2]? Some examples I've found on the internet use Summations and others use Integrals.
 
Yes, they look fine. This is the output of my Mathematica check
upload_2017-10-29_11-23-31.png

Note the two equivalent ways of doing (b).
 

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