# A lemma concerning the distribution of sufficient statistic from exponential family

I understand Lemma 8 in Chapter 1 from Lehmann's Testing Statistical Hypotheses [or Lemma 2.7.2 in Lehmann and Romano] as follows:

If the pdf of an exponential family is $$p_{\theta}(x)=\exp\bigg\{\big(\eta_1^\top(\theta),\eta_2^\top(\theta)\big)\big(T_1(x),T_2(x)\big)^\top-\xi(\theta)\bigg\}h(x),$$ then the pdf of $$T_1$$ is presumably $$g_{\theta}(t)=\exp\{\eta_1^\top(\theta)t-\xi(\theta)\}.$$ However, when I take the joint distribution of $$n$$ iid rv's form normal distribution $$N(\mu, \sigma^2)$$ as an example, where $$\eta(\theta)= \left( \frac{\mu}{2\sigma^2},\frac{1}{2\sigma^2} \right)^\top, T(x)= \left( \sum_{k=1}^n X_k,-\sum_{k=1}^n X_k^2 \right)^\top,\xi(\theta)=\frac{n\mu^2}{2\sigma^2}+\frac{n}{2}\log\sigma^2,h(x)=(2\pi)^{-\frac{n}{2}}$$ The lemma yields that the pdf of $$T_1=\sum_{k=1}^n X_k$$ should be

\begin{align} g_\theta(t) & =\exp\left\{\frac{\mu}{\sigma^2}t-\frac{n\mu^2}{2\sigma^2}-\frac{n}{2}\log\sigma^2\right\} \\[6pt] & = \exp\left\{-\frac{(t-\mu)^2}{2n\sigma^2}+\frac{t^2}{2n\sigma^2}-\frac{n}{2}\log\sigma^2\right\} \end{align}

Whereas, since $$X_k$$'s are iid normal distributions, their sum should also be a normal distribution with mean $$n\mu$$ and variance $$n\sigma^2$$, that is:$$g_\theta(t)=\exp\left\{-\frac{(t-\mu)^2}{2n\sigma^2}-\frac{1}{2} \log(n\pi\sigma^2)\right\}$$ Why do the two results contradict? Did I misunderstand the lemma?

• As @Sextus answered below, if you read the lemma more carefully, you would find it asserted the form of the joint distribution of $(T_1, \ldots, T_s)$, rather than asserting the marginal distribution of $T_1$ alone. So you applied the lemma incorrectly. Commented May 21, 2023 at 14:23

The lemma states that that you can express the probability in that way with respect to a specific measure.

You do not have

$$dP(t)=\exp\{\eta_1^\top(\theta)t-\xi(\theta)\} \, d t$$

but instead you need to use

$$dP(t)=\exp\{\eta_1^\top(\theta)t-\xi(\theta)\} \, d \nu_\theta( t)$$

when you use $$\nu(t) = \int e^{-ct^2} \, dt = \frac{\sqrt{\pi} \operatorname{erf}(\sqrt{c}t)}{2\sqrt{c}}$$ then $$d \nu_\theta( t) = e^{-ct^2} \, dt$$ and

$$dP(t)=\exp\{\eta_1^\top(\theta)t-ct^2-\xi(\theta)\} \,dt$$

You will have to figure out the constant $$c$$ but in this way you have a form that relates to the normal distribution.