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    Iterated variance problem




    I've never posted here before, so I apologize in advance for anything I might do that is not according to forum standards or just plain wrong.

    I'm hoping someone could help me solve a problem that I've been struggling with all night.

    We want to look at a circular sampling region with radius X (feet), where X is exponentially distributed with mean value 10 feet. We also want to look at plants of a certain type, that occur in the region with a Poisson process of 0.5 plants per square foot. Y denotes the number of plants in the region.

    We want to find

    (1) E(Y|X=x)
    (2) V(Y|X=x)
    (3) E(Y)
    (4) V(Y)

    where (3) and (4) are to be computed with the help of (1) and (2).


    So far I've come up with:

    E(Y|X)=0.5\pi x^{2}\qquadV(Y|X)=E(Y|X)=0.5\,\pi\,x^{2}

    since Y is Poisson-distributed. The above is what I gather from the problem text. Next,

    E(Y)=E(E(Y|X))=\int_{-\infty}^{\infty} E(Y|X)\cdot p_{X}(x) dx =\int_{0}^{\infty}0.5\,\pi\,x^{2}\cdot p_{X}(x) dx\\=0.5\,\pi\int_{0}^{\infty}x^{2}\cdot p_{X}(x) dx=0.5\,\pi\, E(X^{2})

    (I see now that I could just have used E(0.5\pi x^{2}) and the properties of the expectation to compute this.)

    Since X is exponentially distributed, we get

    M_{X}(t)=\frac{\lambda}{\lambda-t}=\frac{1}{1-10t}\qquadM_{X}'(t)=\frac{10}{(1-10t)^{2}}

    M_{X}''(t)=\frac{200}{(1-10t)^{3}}\qquadE(X^{2})=M_{X}''(0)=\frac{200}{(1-0)^{3}}=200

    So

    E(Y)=0.5\,\pi\, E(X^{2})=0.5\,\pi\,200=100\,\pi\approx314

    Now,

    V(Y)=E[V(Y|X)]+V[E(Y|X)]

    where

    E[V(Y|X)]=E[E(Y|X)]=100\pi

    since for the Possion distribution, \mu=\sigma^{2} (I'm not totally sure this makes sense in this particular problem, but I don't know how else to find V(Y|X).)

    Next,

    V[E(Y|X)]=V(0.5\pi x^{2})=0.25\,\pi^{2}V(X^{2})=0.25\,\pi^{2}[E(X^{4})-E(X^{2})^{2}]

    M_{X}^{(3)}(t)=\frac{3\cdot\ 10\cdot200}{(1-10t)^{4}}=\frac{6000}{(1-10t)^{4}}, \qquad E(X^{3})=M_{X}^{(3)}(0)=\frac{6000}{(1-0)^{4}}=6000
    M_{X}^{(4)}(t)=\frac{4\cdot\ 10\cdot6000}{(1-10t)^{5}}=\frac{240\,000}{(1-10t)^{5}}, \qquad E(X^{4})=M_{X}^{(4)}(0)=\frac{240\,000}{(1-0)^{5}}=240\,000

    Witch gives us
    V(Y)=E[V(Y|X)]+V[E(Y|X)]=100\pi+0.25\pi^{2}[240\,000-200^{2}]\approx493\,794.37

    But this seems totally wrong, as it implies

    \sigma_{Y}\approx707.71

    witch means we must allow for a negative number of flowers (since the mean number of flowers is approximately 314). However, I have no idea how else to solve this problem.

    I hope this isn't too obvious a question to ask here. I'm just starting to learn statistics, so I realize the above might not be correct, but I hope someone will want to help me anyway.

  2. #2
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    Re: Iterated variance problem

    That doesn't mean you must allow for negative number of flowers. Heck you should be able to reason that just because the sd is greater than the mean doesn't directly imply that there must be negative values. Just play around with the parameters of a gamma distribution - you should easily be able to find some combinations that result in a standard deviation greater than the mean.

    When in doubt simulation to verify answers isn't a bad approach either. If you know R here is some code to simulate your situation...
    Code: 
    > n <- 1000000
    > x <- rexp(n, 1/10)
    > mean(x)
    [1] 10.0052
    > sd(x)
    [1] 9.999635
    > y <- rpois(n, (1/2)*pi*x^2)
    > mean(y)
    [1] 314.3088
    > sd(y)
    [1] 701.8034
    Which is close enough to what you got.

    Also keep in mind that the standard deviation is more sensitive to outliers/skew than the mean is. For example here is what happens if we add a single outlier to some N(1,1) data...
    Code: 
    > x <- rnorm(1000, 1, 1)
    > mean(x)
    [1] 1.021419
    > sd(x)
    [1] 1.031951
    > y <- c(x, 100000)
    > mean(y)
    [1] 100.9205
    > sd(y)
    [1] 3160.666
    The mean increased by approximately 100 and the standard deviation increased by approximately 3160.

    P.S. You don't need to worry about how you're posting. You did a great job with formulating your question and showing the work you've done to get there.

  3. The Following User Says Thank You to Dason For This Useful Post:

    jossan (05-27-2016)

  4. #3
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    Re: Iterated variance problem


    You're right, I should have been able to figure that out.

    Thank you so much! I really appreciate the code examples too. I will try simulating the problem next time.

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