I have great difficulty in calculating $\mathrm{E}\left(\sqrt{T_{1} T_{2}}\right)$ where $T_{i}$ is a random variable distributed according to the Birnbaum-Saunders distribution. Are there any suggestions on how to calculate the expression?

The approximation is in Kundu et. al 2010 but when programming in R the approximation diverges, I'm thinking that the error is in the approach and maybe can meet other approach.

  • $\begingroup$ Are the parameters for the $T_i$ the same? Do you need an exact answer or is an asymptotic approximation acceptable? $\endgroup$
    – Glen_b
    Sep 1, 2013 at 23:10
  • $\begingroup$ Thanks @Glen_b for you interest, $(T_{1},T_{2})\sim BS_2(\alpha_1,\alpha_2,\beta_1,\beta_2,\rho)$ (see link), not an exact value but an acceptable approximation would be great. Best regards. $\endgroup$ Sep 1, 2013 at 23:25
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    $\begingroup$ In the Kundu et al 2010 paper you linked, the expression for this expectation is given in page 118. What is the nature of your difficulty with it? $\endgroup$ Sep 2, 2013 at 2:48
  • $\begingroup$ Thanks @AlecosPapadopoulos for you interest. When programming the expression presented in Kundu pag. 118, the expression diverges (see R program that implements ) and therefore I think the expression given on pag. 119 has some error. $\endgroup$ Sep 2, 2013 at 19:51
  • $\begingroup$ Interesting. If you are certain about that, you should contact the authors of the paper -mistakes in complicated mathematical calculations are sometimes inevitable, and it is good when they are found -and corrected. $\endgroup$ Sep 2, 2013 at 19:54

1 Answer 1


Assuming you know how to compute $\text{E}(T_1 T_2)$ and $\text{Var}(T_1 T_2)$, you can use Taylor series expansion to get an approximation (Wikipedia link, also see this example here on CV):

$g(X)= g(\mu+X-\mu) = g(\mu)+g'(\mu) (X-\mu) + \frac{g''(\mu)}{2} (X-\mu)^2 + ...$


\begin{eqnarray} \text{E}(g(X))&=& g(\mu)+g'(\mu) \text{E}(X-\mu) + \frac{g''(\mu)}{2} \text{E}((X-\mu)^2) + ...\\ &=& g(\mu)+ 0 + \frac{g''(\mu)}{2} \text{Var}(X) + ... \end{eqnarray}


$\text{E}(\sqrt{T_1 T_2}) \approx \sqrt{\text{E}(T_1 T_2)} -\frac{1}{8}\text{E}(T_1 T_2)^{-\frac{3}{2}} \text{Var}(T_1 T_2)$

(assuming I made no errors)

You can carry the expansion out further, but usually for expectations it is only taken out to the variance term.

  • $\begingroup$ Thanks very you much @Glen_b. I will review their expression. $\endgroup$ Sep 1, 2013 at 23:50
  • $\begingroup$ The difficulty now is to calculate $E(T_{1}T_{2})$, the paper of link Kundu et. the 2010 uses an approach that when I program in R diverges, so I think something is wrong. $\endgroup$ Sep 2, 2013 at 0:20
  • $\begingroup$ $\text{E}(T_1T_2) = \text{Cov}(T_1,T_2) + \text{E}(T_1)\text{E}(T_2)= \rho \sigma_1 \sigma_2 + \text{E}(T_1)\text{E}(T_2)$ $\endgroup$
    – Glen_b
    Sep 2, 2013 at 0:31
  • $\begingroup$ Thanks @Glen_b for you help, also share a formula more see link $Var(T_{1}T{2})=Cov(T_{1}^2,T_{2}^2) + (Var(T_{1}) + E(T_{1})^2)(Var(T_{2}) + E(T_{2})^2) - (Cov(T_{1},T_{2}) + E(T_{1})E(T_{2}))^2$ $\endgroup$ Sep 2, 2013 at 2:42
  • $\begingroup$ Good. If you were happy with the first formula I was planning to go dig up the corresponding formula for the variance. Looks like you're able to at least compute the approximation. If the distribution isn't fairly 'tight' around the mean (small coefficient of variation, in the sense of zero being more than a couple of sds from the mean, for example), then the approximation often isn't so great. $\endgroup$
    – Glen_b
    Sep 2, 2013 at 4:27

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