Let L be the random variable for the length of time, in years, that a person will remember an actuarial statistic. For a certain popula- tion, L is exponentially distributed with mean 1/Y, where Y has a gamma distribution with a = 4.5 and 3 = 4. Find the variance of L.

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
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Chapter10: Statistics
Section10.5: Comparing Sets Of Data
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Let L be the random variable for the length
of time, in years, that a person will remember
an actuarial statistic. For a certain popula-
tion, L is exponentially distributed with mean
1/Y, where Y has a gamma distribution with
a = 4.5 and 3 = 4. Find the variance of L.
Transcribed Image Text:Let L be the random variable for the length of time, in years, that a person will remember an actuarial statistic. For a certain popula- tion, L is exponentially distributed with mean 1/Y, where Y has a gamma distribution with a = 4.5 and 3 = 4. Find the variance of L.
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No, this solution is not correct. The variance of an exponentially distributed random variable is the square of its mean, so Var(L) = (1/Y)^2. However, the given solution incorrectly multiplies this by Var(Y) to obtain an incorrect expression for Var(L).

The correct approach to find Var(L) involves using the law of total variance. The law of total variance states that Var(L) = E[Var(L|Y)] + Var(E[L|Y]). We already know that Var(L|Y) = (1/Y)^2, so E[Var(L|Y)] = E[(1/Y)^2]. We also know that E[L|Y] = 1/Y, so Var(E[L|Y]) = Var(1/Y).

To find E[(1/Y)^2] and Var(1/Y), we need to use the formulas for the expected value and variance of a function of a random variable. These involve integrating the function multiplied by the probability density function of the random variable.

Once we correctly compute E[(1/Y)^2] and Var(1/Y), we can plug the correct values into the law of total variance to find the correct value for Var(L).

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Follow-up Question

But isn't the mean of the gamma distribution aplha (beta)

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Follow-up Question

but isn't the expectation alpha times beta rather than alpha divided by beta?

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