# Variance of sum calculation in example illustrating completeness for minimally sufficient statistic

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.

The notation for the normal distribution is $$N(\mu, \sigma^2)$$. So if you have $$\sum Y_i \sim N(n\mu, na^2\mu^2)$$, the variance is equal to $$na^2\mu^2$$ by definition.