Geometric Random Variable probability problem

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Discussion Overview

The discussion revolves around a homework problem involving a geometric random variable with a probability parameter of p = 0.1. Participants are tasked with calculating various probabilities, expectations, and variances related to the random variable, including specific values and conditional expectations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents calculations for several parts of the problem, including cumulative distribution function values and conditional probabilities.
  • Another participant questions the definition of expectation value, suggesting it is the average.
  • Some participants express agreement with the calculations provided, noting that excessive decimal precision may be distracting.
  • A participant inquires about alternative methods for calculating expected values, mentioning that some resources use summations while others use integrals.
  • One participant confirms the correctness of the calculations using Mathematica, indicating that there are equivalent methods for certain calculations.

Areas of Agreement / Disagreement

Participants generally agree on the correctness of the calculations for parts a, b, and c, but there is no consensus on the best method for calculating expected values or the appropriateness of decimal precision in answers.

Contextual Notes

Some calculations depend on the interpretation of the geometric random variable and the definitions of expectation and variance. There are unresolved questions regarding the methods for obtaining expected values and the level of precision required in answers.

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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