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Mean of Exponential RVs

  1. Apr 25, 2010 #1
    I forgot how to derive the mean of exponential random variables follow the chi square distribution with degree freedom 2n. Don't know where I got it wrong. Anyone have a clue how to do it?

    Thanks
     
  2. jcsd
  3. Apr 26, 2010 #2
    [tex]\mu = \int_0^{\infty}t f(t)dt[/tex] where [tex]f(t)[/tex] is the pdf.
     
  4. Apr 26, 2010 #3
    I want the distribution of the sample mean, which should has a chi square distribution. I know how to do the expectation of the sample mean, which is u.

    Not quite what I am asking for. Thanks for the reply
     
  5. Apr 26, 2010 #4

    statdad

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    Homework Helper

    try calculating the moment-generating function of the mean (it exists, since all moments of the exponential distribution exist). Note that for ANY distribution, where all the following integrals exist, the moment generating function of the sample mean is

    [tex]
    m_{\overline x} (s) = E[e^{s \frac 1 n \sum x}] = E[\prod e^{(\frac s n)x}] = \prod{E[e^{(\frac s n)x}]}
    [/tex]

    and each factor in the final product is calculated based on the exponential distribution.
     
  6. Apr 26, 2010 #5
    I don't get it by doing MGF. Sum of exponential RVs is a Gamma RV. Chi square is the special case of Gamma with parameter is 1/2. I don't know how to from a Gamma by taking average to get the Chi.
     
  7. Apr 28, 2010 #6
    Well the exponential distribution is a special case of the gamma dist. where k=1. In addition, when lambda=1/2, I believe the distribution of the sample means asymptotically approaches a chi square dist. with a mean of 2 (2 degrees of freedom). If so, what exactly is your question?
     
  8. Apr 29, 2010 #7
    If you're asking about a hyper-exponential dist.; it is still a one parameter dist., not a higher form (k>1) of the gamma dist.
     
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