$Y_1 + \dots + Y_n \sim N(n\mu, n\sigma^2)$ implies that $\bar{Y} \sim N(\mu, \sigma^2/n)$?

I have this example of sufficiency:

Let $$Y_1, \dots, Y_n$$ be i.i.d. $$N(\mu, \sigma^2)$$. Note that $$\sum_{i = 1}^n (y_i - \mu)^2 = \sum_{i = 1}^n (y_i - \bar{y})^2 + n(\bar{y} - \mu)^2$$. Hence

\begin{align} L(\mu, \sigma; \mathbf{y}) &= \prod_{i = 1}^n \dfrac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{1}{2\sigma^2}(y_i - \mu)^2} \\ &= \dfrac{1}{(2\pi \sigma^2)^{n/2}}e^{-\frac{1}{2\sigma^2}\sum_{i = 1}^n (y_i - \bar{y})^2}e^{-\frac{1}{2\sigma^2}n(\bar{y} - \mu)^2} \end{align}

From Theorem 1, it follows that where $$T(\mathbf{Y}) = (\bar{Y}, \sum_{i = 1}^n (Y_i - \bar{Y})^2)$$ is a sufficient statistic for $$(\mu, \sigma)$$.

It then says the following:

We now show that $$\bar{Y} \sim N(\mu, \frac{\sigma^2}n)$$.

It is clear that

$$Y_1 + \dots + Y_n \sim N(n\mu, n\sigma^2)$$

and so

$$\bar{Y} \sim N\left( \mu, \frac{\sigma^2}n \right)$$

Theorem 1 is presented as follows:

A statistic $$T(\mathbf{Y})$$ is sufficient for $$\theta$$ if, and only if, for all $$\theta \in \theta$$

$$L(\theta; \mathbf{y}) = g(T(\mathbf{y}), \theta) \times h(\mathbf{y})$$

where the function $$g(\cdot)$$ depends on $$\theta$$ and the statistic $$T(\mathbf{Y})$$, while the function $$h(\cdot)$$ does not contain $$\theta$$.

Theorem $$1$$ implies that if the likelihood $$L(\theta; \mathbf{y})$$ depends on the data only through $$T(\mathbf{y})$$, $$T(\mathbf{Y})$$ is a sufficient statistic for $$\theta$$ and $$h(\mathbf{y}) \equiv 1$$.

How does $$Y_1 + \dots + Y_n \sim N(n\mu, n\sigma^2)$$ imply that $$\bar{Y} \sim N(\mu, \frac{\sigma^2}n)$$?

We know that if $$A \sim N(\mu, \sigma^2)$$, then $$kA \sim N(k\mu, k^2 \sigma^2)$$.

Hence in this case, $$A = \sum_{i=1}^n Y_i$$ and $$k =\frac1n$$ since we have

$$\bar{Y}=\frac{\sum_{i=1}^n Y_i}{n}$$

Edit:

We have $$\sum_{i=1}^nY_i \sim N(n \mu, n \sigma^2)$$

Hence

$$\frac{\sum_{i=1}^nY_i}{n}\sim N(\frac{n\mu}{n},\frac{n\sigma^2}{n^2})$$

• But these notes had $\bar{Y} \sim N(\mu, \sigma^2/n)$. If what you're saying is correct, then shouldn't we have $\bar{Y} \sim N(\mu/n, \sigma^2/n^2)$? Commented Mar 14, 2020 at 6:01
• I have edited my answer, we started from $N(n\mu, n\sigma^2)$, hence when we divide the random variable by $n$, the mean is $\mu$ and the variance is $\frac{\sigma^2}{n}$. Commented Mar 14, 2020 at 6:05
• Why does the division by $n$ here divide the mean by $n$ but the variance by $n^2$? I must be missing some understanding about this. Commented Mar 14, 2020 at 6:13
• $Var(X)=E[X^2]-(E[X])^2$, hence $$Var(kX)=E[(kX)^2]-(E(kX))^2=E[k^2X^2]-k^2(E[X))^2=k^2(E[X^2]-E[X]^2)=k^2Var(X)$$ Commented Mar 14, 2020 at 6:18
• Yes, interpret $X = \sum_{i=1}^n Y_i$ and the transformation is to convert it to $\frac{X}{n}$ Commented Mar 14, 2020 at 8:02