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Math Help - Weibull distribution

  1. #1
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    Weibull distribution

    Having trouble on this question:
    The random variable Y with a density function given by
    f(y)=[[(m)y^(m-1)]/(a)]*e^[(-y^(m))/(a)] where y is between 0 and infinity and a,m are greater than zero
    is said to have a Weibull distribution. Find the mean and variance for a Weibull distributed random variable with m=2.

    I can find the answers on the Internet somewhere, but how do you actually derive them?

    Thanks for any help.
    Jack
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  2. #2
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    OK. I found the mean, but could someone please help me with the variance? I'm stuck.

    Thanks in anticipation.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by JackStolerman View Post
    Having trouble on this question:
    The random variable Y with a density function given by
    f(y)=[[(m)y^(m-1)]/(a)]*e^[(-y^(m))/(a)] where y is between 0 and infinity and a,m are greater than zero
    is said to have a Weibull distribution. Find the mean and variance for a Weibull distributed random variable with m=2.

    I can find the answers on the Internet somewhere, but how do you actually derive them?

    Thanks for any help.
    Jack
    You have density:

    <br />
f_{m,a}(y) = \frac{m y^{m-1}}{a}e^{-y^m/a}<br />

    where m,\ a >0 and y \in (0, \infty)

    Then the mean is:

    <br />
\mu = \int_0^{\infty} y\ f_{m,a}(y)\ dy<br />

    which you say you can do.

    The variance is given by:

    <br />
\sigma^2 = \int_0^{\infty} (y-\mu)^2 \ f_{m,a}(y)\ dy = <br />
\int_0^{\infty} y^2 \ f_{m,a}(y)\ dy -\mu^2<br />

    Now the integral on the right in this last equation should be do-able
    using the same technique you used to evaluate the mean.

    RonL
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  4. #4
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    Ansan, South Korea
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    Re: Weibull distribution

    Consider a Weibull distribution with the following pdf:
    f(x)= (Ɵ2/ Ɵ1)*(x/ Ɵ1)^( Ɵ2-1)*exp{-( x/ Ɵ1)^ Ɵ2} For x>0, Ɵ1>0, Ɵ2>0
    (a) if now Y=( x/ Ɵ1)^ Ɵ2. Show that Y follows an exponential distribution with a mean of 1.
    Explain how to
    (b) Explain how to generate a random variate from this Weibull distribution based on Y.
    (c) Use part (a) to generate 500 random variates from this Weibull distribution when (Ɵ1, Ɵ2)=(3,3) and draw a histogram.
    (d) Compare the histogram in part (c) with the actual pdf of Weibull (3,3).
    (e) Now assume that this Weibull distribution is truncated at points (a,b) = (1,10). Generate 500 random variates from this trunacated Weibull distribution and compute the average.
    (f) Compute the (theoretical expected value of this truncated population. Is the average in part(e) close to this expected value?
    I am New in statistics and got this question as my homework. I tried my best but not able to solve this. Please help me both in Theoretical as well as Matlab coding for the above question
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