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


1 Answer 1


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

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


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


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

  • $\begingroup$ But these notes had $\bar{Y} \sim N(\mu, \sigma^2/n)$. If what you're saying is correct, then shouldn't we have $\bar{Y} \sim N(\mu/n, \sigma^2/n^2)$? $\endgroup$ Commented Mar 14, 2020 at 6:01
  • $\begingroup$ 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 $\frac{\sigma^2}{n}$. $\endgroup$ Commented Mar 14, 2020 at 6:05
  • $\begingroup$ 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. $\endgroup$ Commented Mar 14, 2020 at 6:13
  • $\begingroup$ $Var(X)=E[X^2]-(E[X])^2$, hence $$Var(kX)=E[(kX)^2]-(E(kX))^2=E[k^2X^2]-k^2(E[X))^2=k^2(E[X^2]-E[X]^2)=k^2Var(X)$$ $\endgroup$ Commented Mar 14, 2020 at 6:18
  • 1
    $\begingroup$ Yes, interpret $X = \sum_{i=1}^n Y_i$ and the transformation is to convert it to $\frac{X}{n}$ $\endgroup$ Commented Mar 14, 2020 at 8:02

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