I have an example where it is said that $$\sum_{i = 1}^n Y_i \sim N(n \mu, n a^2 \mu^2)$$ and
$$\begin{align} E \left[ \left( \sum_{i = 1}^n Y_i \right)^2 \right] &= \text{Var} \left( \sum_{i = 1}^n Y_i \right) + \left[ E \left( \sum_{i = 1}^n \right) \right]^2 \\ &= na^2\mu^2 + n^2 \mu^2 = n(a^2 + n)\mu^2\end{align}$$
It seems that this implies that $\text{Var} \left( \sum_{i = 1}^n Y_i \right) = na^2\mu^2$, which, if I'm not mistaken, is not in accordance with the properties of variance (there should be a covariance in there somewhere?).
The full example is for the purpose of illustrating completeness in regards to minimally sufficient statistics:
Example 10
Let $Y_1, \dots, Y_n$ be i.i.d. $N(\mu, a^2 \mu^2)$ and $a > 0$ is known.
Such a model describes, for example, a series of measurements on an object using a measuring device whose accuracy is relative to the true (unknown) measure of the object rather than being constant.
Show that $T(\mathbf{Y}) = (\sum_{i = 1}^n Y_i, \sum_{i = 1}^n Y_i^2)$ is
- a minimally sufficient statistic for $\mu$ and
- not a complete statistic.
Take $\mathbf{y}_1$ and $\mathbf{y}_2$ be two sets of sample. The ratio of the likelihood is
$$\begin{align} \dfrac{L(\mu; \mathbf{y}_1)}{L(\mu; \mathbf{y}_2)} = \dfrac{\dfrac{1}{(\sqrt{2 \pi a^2 \mu^2})^n} e^{-\dfrac{1}{2 a^2 \mu^2}\sum_{i = 1}^n (y_{1i} - \mu)^2}}{\dfrac{1}{(\sqrt{2 \pi a^2 \mu^2})^n} e^{-\dfrac{1}{2 a^2 \mu^2}\sum_{i = 1}^n (y_{2i} - \mu)^2}} &= e^{-\dfrac{\sum_{i = 1}^n (y_{1i} - \mu)^2 - \sum_{i = 1}^n (y_{2i} - \mu)^2}{2a^2\mu^2}} \\ &= e^{-\dfrac{\sum_{i = 1}^n y^2_{1i} - 2 \sum_{i = 1}^n y_{1i} \mu - \sum_{i = 1}^n y^2_{2i} + 2 \sum_{i = 1}^n y_{2i} \mu}{2 a^2 \mu^2}} \end{align}$$
The ratio doesn't depend on $\mu$ if $\sum_{i = 1}^n y^2_{1i} = \sum_{i = 1}^n y^2_{2i}$ and $\sum_{i = 1}^n y_{1i} = \sum_{i = 1}^n y_{2i}$. So $T(\mathbf{Y})$ is a minimal sufficient statistic.
It is clear that $\sum_{i = 1}^n Y_i \sim N(n \mu, n a^2 \mu^2)$. We have
$$\begin{align} E \left[ \left( \sum_{i = 1}^n Y_i \right)^2 \right] &= \text{Var} \left( \sum_{i = 1}^n Y_i \right) + \left[ E \left( \sum_{i = 1}^n \right) \right]^2 \\ &= na^2\mu^2 + n^2 \mu^2 = n(a^2 + n)\mu^2\end{align}$$
$$\begin{align} E \left[ \sum_{i = 1}^n Y_i^2 \right] &= \sum_{i = 1}^n E[Y^2_i] \\ &= \sum_{i = 1}^n \left( \text{Var}(Y_i) + E[Y_i]^2 \right) \\ &= n(a^2 + 1)\mu^2 \end{align}$$
This implies that
$$E \left[ \dfrac{a^2 + n}{a^2 + 1} \sum_{i = 1}^n Y_i^2 - \left( \sum_{i = 1}^n Y_i \right)^2 \right] = 0$$
I would greatly appreciate it if people would please take the time to clarify how $\text{Var} \left( \sum_{i = 1}^n Y_i \right)$ is computed in this example.
