Calculating Variance of a Sum of Independent Random Variables

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SUMMARY

The discussion focuses on calculating the variance of a sum of independent random variables, specifically in the context of donations collected through two Poisson-distributed call options. The variables X and Y represent the number of calls made to two different numbers, with X following a Poisson distribution of 1500 calls/minute and Y at 3750 calls/minute. The expected total donation, represented as Z = 9.9X + 0.5Y, is calculated to be 2,007,000. The challenge lies in determining the standard deviation of Z, as the standard method using \sqrt{μ} does not yield the correct result.

PREREQUISITES
  • Understanding of Poisson distribution, specifically Poisson(180000) and Poisson(450000).
  • Knowledge of expectation and variance in probability theory.
  • Familiarity with the formula for the variance of a linear combination of independent random variables.
  • Basic proficiency in statistical approximation techniques, particularly normal approximation.
NEXT STEPS
  • Study the variance formula for independent random variables: Var(aX + bY) = a²Var(X) + b²Var(Y).
  • Learn about the Central Limit Theorem and its application to Poisson distributions.
  • Explore methods for calculating standard deviation in the context of combined distributions.
  • Investigate the implications of using normal approximation for Poisson-distributed variables.
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Students and professionals in statistics, data analysis, and probability theory, particularly those working with Poisson processes and variance calculations in random variables.

Gauss M.D.
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Homework Statement



During a two hour window, people are given the option of calling number X, donating $9.90, or number Y, donating $0.50.

X is Poisson distributed with 1500 calls/minute. Y is Poisson with 3750 calls/minute.

What is the probability that more than $2,000,000 is raised?

Homework Equations





The Attempt at a Solution



X = number of calls to number X in 120 minutes = Po(120*1500) = Po(180000)

Y = number of calls to number Y in 120 minutes = Po(120*3750) = Po(450000)

Let Z = 9.9X + 0.5Y. We're looking for P(Z > 2,000,000).

Z should have an expectation of 1,782,000 + 225,000 = 2,007,000.

Now I want to approximate Z = Po(μ) with Z ≈ N(μ,σ). But I can't work out the standard deviation for Z. Using \sqrt{μ} doesn't give me the right answer.
 
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Gauss M.D. said:

Homework Statement



During a two hour window, people are given the option of calling number X, donating $9.90, or number Y, donating $0.50.

X is Poisson distributed with 1500 calls/minute. Y is Poisson with 3750 calls/minute.

What is the probability that more than $2,000,000 is raised?

Homework Equations





The Attempt at a Solution



X = number of calls to number X in 120 minutes = Po(120*1500) = Po(180000)

Y = number of calls to number Y in 120 minutes = Po(120*3750) = Po(450000)

Let Z = 9.9X + 0.5Y. We're looking for P(Z > 2,000,000).

Z should have an expectation of 1,782,000 + 225,000 = 2,007,000.

Now I want to approximate Z = Po(μ) with Z ≈ N(μ,σ). But I can't work out the standard deviation for Z. Using \sqrt{μ} doesn't give me the right answer.

I would bet that you have already seen how to do it, but may have forgotten. So, if X and Y are independent random variables and a, b are numbers, how does the variance of Z = a*X + b*Y relate to Var(X) and Var(Y)? There is a standard formula. It is used over and over and over again, so you should get to know it if you don't already or have forgotten it.
 

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