Assume a scalar random variable $X$ belongs to a vector-parameter exponential family with p.d.f.

$$ f_X(x|\boldsymbol \theta) = h(x) \exp\left(\sum_{i=1}^s \eta_i({\boldsymbol \theta}) T_i(x) - A({\boldsymbol \theta}) \right) $$

where ${\boldsymbol \theta} = \left(\theta_1, \theta_2, \cdots, \theta_s \right )^T$ is the parameter vector and $\mathbf{T}(x)= \left(T_1(x), T_2(x), \cdots,T_s(x) \right)^T$ is the joint sufficient statistic.

It can be show that the mean and the variance for each $T_i(x)$ exist. However, do the mean and the variance for $X$ (i.e. $E(X)$ and $Var(X)$) always exist as well? If not, is there an example of an exponential family distribution of this form whose mean and variable do not exist?

Thank you.


Taking $s=1$, $h(x)=1$, $\eta_1(\theta)=\theta$, and $T_1(x)=\log(|x|+1)$ gives $A(\theta)=\log\left(-2/(1+\theta)\right)$ provided $\theta \lt -1$, producing

$$f_X(x|\theta) = \exp\left(\theta\log(|x|+1) - \log\left(\frac{-2}{1+\theta}\right)\right) = -\frac{1+\theta}{2}(1+|x|)^\theta. $$


Graphs of $f_X(\ |\theta)$ are shown for $\theta=-3/2, -2, -3$ (in blue, red, and gold, respectively).

Clearly the absolute moments of weights $\alpha=-1-\theta$ or greater do not exist, because the integrand $|x|^\alpha f_X(x|\theta)$, which is asymptotically proportional to $|x|^{\alpha+\theta}$, will produce a convergent integral at the limits $\pm\infty$ if and only if $\alpha+\theta\lt -1$. In particular, when $-2 \le \theta \lt -1,$ this distribution does not even have a mean (and certainly not a variance).

  • $\begingroup$ I do not understand the condition $\theta < -1$. Do you mean $\theta > -1$? When $\theta < -1$, $A(\theta)$ is not defined and $f_X(x|\theta)$ is negative and cannot be a p.d.f. Please let me know what I missed. Thanks. $\endgroup$ – Wei Mar 20 '14 at 22:27
  • $\begingroup$ I apologize, because a minus sign was omitted in the calculation of $A$. I have replaced it in the formulas. I really do mean $\theta\lt -1$. $\endgroup$ – whuber Mar 21 '14 at 1:00
  • $\begingroup$ Thank you for the example. I agree about the moments of $|x|$. How about the moments of $x$ itself? For example, when $-2<\theta <-1$ in your example above, does $E(x)$ exist? $\endgroup$ – Wei Mar 30 '14 at 12:21
  • 1
    $\begingroup$ Because the Lebesgue integral is defined in terms of the positive and negative parts of the integrand, the moments of $x$ exist if and only if the moments of $|x|$ exist. $\endgroup$ – whuber Mar 30 '14 at 17:32
  • $\begingroup$ @Wei: $\mathrm{E}\{g(X)\}$ exists only if $\mathrm{E}\{\, \left| g(X)\, \right| \} < \infty$. Without this restriction,the expectation is not uniquely defined for some CDFs. $\endgroup$ – Dennis Jul 12 '14 at 0:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.