mean of an absolute Student’s t

Having (rather foolishly) involved myself into providing an answer for Cross Validated: “Can the standard deviation of non-negative data exceed the mean?“, I ended up having to derive the mean of the absolute value of a Student’s variate X.  (Well, not really, but then I did.) I think the following is correct:

\mathbb{E}[|X|]=\mu(2\text{P}_{\mu,\nu}(X>0)-1)\qquad\qquad\\ \qquad\qquad+2 f(0,\nu)\dfrac{\nu}{\nu-1}\left[1+\mu^2/\nu\right]^{-(\nu-1)/2}

where f(x,\nu) is the density of the standard Student’s distribution. (I also checked by simulation. The derivation is there. And now fully arXived. Even though it is most likely already written somewhere else. See, for instance, Psarakis and Panaretos who studied the case of the absolute centred t rv, called the folded t rv.)

4 Responses to “mean of an absolute Student’s t”

  1. Hello,
    a recent article that may interest you:
    Some results on the truncated multivariate t distribution
    Journal of Statistical Planning and Inference Vol 142(1) Pages 25-40
    Hsiu J. Ho, Tsung-I. Lin, Hsuan-Yu Chen, Wan-Lun Wang
    “We provide explicit matrix expressions for the first two moments of the truncated multivariate t (TMVT) distribution. We apply the slice sampling to generate random variates from the TMVT distribution by introducing auxiliary variables. We describe an algorithm to draw TMVT samples from a series of full conditionals that are of uniform distributions. Practical applications are given to demonstrate the effectiveness and importance of our methodology.”

  2. Often when dealing with such questions, simple discrete distributions (like the Bernoulli or sometimes adding a third point) are are a quick route to examples and counterexamples. In this case the Bernoulli with p<0.5 is sufficient to answer the question since it has standard deviation greater than its mean. So a small data set with a 1 and several 0's should be sufficient, even if the n-1 denominator is used in the sample calculation.

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