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    Equation involving uniformly distributed variable




    Let U be a random variable uniformly distributed on [0, 1], and let X be a dependent variable such that X^U = a for some number a (X^U means X to the power of U). Is there any way to estimate the expected value of X?

    If X were a number x, I could integrate this equality (multiplied by the pdf of the uniform distribution) from u = 0 to u = 1 to get an equation for x but perhaps the expected value of X need not satisfy this equation …

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    Re: Equation involving uniformly distributed variable

    X^U = a \Rightarrow X = a^{\frac {1} {U}}

    When a \in (0, 1], the expectation E[X] exist.

    However, you will need to find the expectation by simulation/numerical integration.

    For example, when a = \frac {1} {2}, using the Simpson's rule,

    n<-1000000
    sum<-0
    for (i in 1:n) {
    sum<-sum+(a^(n/(i-1))+4*a^(2*n/(2*i-1))+a^(n/i))
    }
    sum/6/n
    we have E[X] \approx 0.2375252

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    Re: Equation involving uniformly distributed variable

    Thank you very much, BGM. Why do you note that a should be between 0 and 1? Do you mean that E[X] does not exist if a > 1?

    While I can certainly perform numeric integration to come up with an approximation, I still wonder if an analytic estimate exists for the E[X]. If I assume that X is a number x, then, integrating the equality from u = 0 to u = 1 gives us this equation: (x – 1)/ln(x) = a. For a = ˝ this gives us x = 0.20315 which is close to the result you got using the Simpson’s rule.

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    Re: Equation involving uniformly distributed variable

    The expectation does not exist if a > 1. The integral diverges.

    We may try the Taylor expansion around E[ U]:

    Consider g(u) \triangleq a^{\frac {1} {u}}

    \Rightarrow E[X] = E[g(U)]

    \approx E\left[g(E[ U]) + (U - E[ U])g'(E[ U])+ \frac {1} {2} (U - E[ U])^2g''(E[ U])\right]

    = g(E[ U]) + \frac {1} {2} g''(E[ U])Var[ U]

    In our case,

    E[ U] = \frac {1} {2}, Var[ U] = \frac {1} {12},g''(u) = \frac {a^{\frac {1} {u}}\ln a} {u^4} (2u + \ln a)

    Substitute inside, we get

    E[X] \approx a^2 + \frac {1} {2} \times 16 a^2 (\ln a) (1 + \ln a) \times \frac {1} {12} = \left[1 + \frac {2} {3} (\ln a + (\ln a)^2) \right] a^2

    Put a = \frac {1} {2}, we get the approximation 0.2145510
    Last edited by BGM; 03-31-2011 at 01:51 PM. Reason: Correct for the typo

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    Re: Equation involving uniformly distributed variable

    Thank you again. This is very helpful. There is a small typo in your calculation of g''(u): one should replace ln u by ln a.

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    Re: Equation involving uniformly distributed variable


    Thanks for the reminder.

    Just one more thing: you can always add more terms in the expansion if you
    are not satisfy with the approximation.

    For instance, since the uniform distribution is symmetric, the next term is
    in order 4.

    g^{(4)}(u) = \frac {a^{\frac {1} {u}}\ln a} {u^8} \left[24u^3 + 36 \ln a u^2 + 12(\ln a)^2 u + (\ln a)^3 \right]

    Hence the next term is

    \frac {1} {4!} E[(U - E[ U])^4] g^{(4)} (E[ U])

    = \frac {1} {24} \times \frac {1} {80} \times 2^8 a^2\ln a [3 + 9 \ln a + 6(\ln a)^2 + (\ln a)^3]

    = \frac {2} {15} \times \ln a [3 + 9 \ln a + 6(\ln a)^2 + (\ln a)^3] \times a^2

    And hence this order 4 approximation becomes

    E[X] \approx \left[1 + \frac {2} {15} (8\ln a + 14 (\ln a)^2 + 6(\ln a)^3 + (\ln a)^4)  \right] a^2

    This time putting a = \frac {1} {2} gives 0.2304617

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