1

I have a list of means and variances. I want to average them both. For mean I simply put: $$\mu = \frac{\mu_1+\mu_2+\cdots+\mu_n}{n}$$

However, for the variance, I can't figure out to average $n$ variances. For only two I found this answer here

$$\sigma^2 = \frac{1}{2} \left(\sigma^2_1 + \sigma^2_2\right) + \left(\frac{\mu_1-\mu_2}{2}\right)^2$$

But how about $n$ variances? thank you

Michael Hardy
  • 1
Meryem Janati
  • 19
  • $$ \text{It's } \frac{\sigma_1^2 + \cdots + \sigma_n^2} n + \frac{(\mu_1-\overline \mu)^2 + \cdots +(\mu_n-\overline \mu)^2} n \text{ where } \overline\mu = \frac{\mu_1 + \cdots + \mu_n} n. $$ – Michael Hardy Nov 16 '20 at 03:29

1 Answers1

0

I will surmise that what you mean is the variance of the mean of $n$ random variables, and each has $k$ possible values.

$$\text{It's } \frac{\sigma_1^2 + \cdots + \sigma_n^2} n + \frac{(\mu_1-\overline \mu)^2 + \cdots +(\mu_n-\overline \mu)^2} n \text{ where } \overline\mu = \frac{\mu_1 + \cdots + \mu_n} n.$$

Let $x_{i,1},\ldots ,x_{i,k}$ be the possible values of the $i$ random variable for $i=1,\ldots,n.$ Then \begin{align} \require{cancel} & \mu_i = \frac{x_{i,1} + \cdots + x_{i,k}} k \\[10pt] & \sigma_i^2 = \frac{(x_{i,1} - \mu_i)^2 + \cdots + (x_{i,k} - \mu_i)^2} k \\[10pt] & \sigma_i^2 + (\mu_i-\overline \mu)^2 = \frac 1 k \sum_{j=1}^k \big( (x_{i,j} - \mu_i)^2 + (\mu_i - \overline\mu)^2 \big) \\[10pt] = {} & \frac 1 k \sum_{j=1}^k \big( (x_{i,j} - \mu_i)^2 - 2(x_{i,j} - \mu_i) (\mu_i - \overline\mu) + (\mu_i - \overline\mu)^2 \big) \tag 1 \\ & \textbf{Why? See below.} \end{align} This last equality is true because \begin{align} & \sum_{j=1}^k \big( (x_{i,j} - \mu_i) (\mu_i - \overline\mu) \big) \\[8pt] = {} & (\mu_i-\overline \mu)\sum_{j=1}^k (x_{i,j} - \mu_i)\\ & \text{since $\mu_i-\overline\mu$ does not change as $j$ goes from $1$ to $k$} \\[10pt] = {} & (\mu_i-\overline\mu)\cdot0. \end{align}

Then line $(1)$ above becomes \begin{align} & \frac 1 k \sum_{j=1}^k \big( (x_{i,j} - \mu_i)^2 - 2(x_{i,j} - \mu_i) (\mu_i - \overline\mu) + (\mu_i - \overline\mu)^2 \big) \\[8pt] = {} & \frac 1 k \sum_{j=1}^k \big( (x_{i,j} - \mu_i) - (\mu_i - \overline\mu) \big)^2 = \frac 1 k \sum_{j=1}^k (x_{i,j} - \overline \mu)^2. \end{align}

Michael Hardy
  • 1