Geometric Random Variable probability problem

In summary, we found the values for a geometric random variable X with p = 0.1. Using the formula for cumulative distribution function, we calculated F_X(5) = 0.40951 and Pr(5 < X \leq 11) = 0.2766794039. For Pr(X=7|5<X\leq11), we used the formula for probability mass function to get 0.0531441, and then calculated the conditional probability as 0.192078. For E(X|3<X\leq11), we used the formula for expected value to get 6.9533987498770702912469710479707. For E(X^2|3<X
  • #1
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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  • #2
How is an expectation value defined?
 
  • #3
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?
 
  • #4
I get the same thing. Note that you would typically not need to quote so many decimals, it is essentially just distracting.
 
  • #5
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.
 
  • #6
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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1. What is a geometric random variable?

A geometric random variable is a discrete random variable that represents the number of trials needed to achieve a success in a sequence of independent trials with a constant probability of success.

2. How is the probability of a geometric random variable calculated?

The probability of a geometric random variable can be calculated using the formula P(X = k) = (1-p)^(k-1) * p, where p is the probability of success and k is the number of trials.

3. What is the expected value of a geometric random variable?

The expected value of a geometric random variable is equal to 1/p, where p is the probability of success.

4. Can a geometric random variable have a value of 0?

No, a geometric random variable cannot have a value of 0 since it represents the number of trials needed to achieve a success.

5. How is a geometric random variable used in real life?

A geometric random variable can be used to model situations where there is a constant probability of success in a series of independent trials, such as the number of attempts needed to make a successful basketball shot or the number of tries to win a game of chance.

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