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.


1 Answer 1


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
    Commented Mar 20, 2014 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
    Commented Mar 21, 2014 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
    Commented Mar 30, 2014 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
    Commented Mar 30, 2014 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
    Commented Jul 12, 2014 at 0:23

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