Let $x_1,\ldots,x_n$ be any numbers and $\bar{x} = (x_1+ \ldots + x_n)/n$. Then I need to prove that $$min_a \sum_{i=1}^n (x_i - a)^2 = \sum_{i=1}^n (x_i - \bar{x})^2$$ 

I do not need the answer but I need an visual/intuitive understanding of the question. I do not understand what min$_a$ mean. Actually I'm kind of struggling to understand "minimization concepts" i.e. the value $a$ is not defined or anything so what does minimization of $a$ mean ? It would be great if someone can explain the concept in general too. Thanks.

I got the answer perfectly for what I asked. I tried to solve the problem as well. I got stuck but here is my try.

$$\sum_{i=1}^n (x_i - a)^2 = \sum_{i=1}^n (x_i - \bar{x} + \bar{x} - a)^2$$
$$= \sum_{i=1}^n (x_i - \bar{x})^2 + 2 \sum_{i=1}^n (x_i - \bar{x})(x_i - a) + \sum_{i=1}^n(x_i - a)^2 $$ ...

Can anyone help me after this ? Thanks again.

These are the likelihoods of the corresponding set of $10$ coin tosses having been produced by the two coins (using the current estimate for their biases) normalized to add up to $1$. The estimated probability of $k$ out of $10$ tosses of coin $i$ ($i\in\{A,B\}$) yielding heads is

$$p_i(k)=\left({10\atop k}\right) \theta_i^k (1-\theta_i)^{10-k}\;.$$

The binomial coefficient is the same for both coins, so it drops out in the normalization, and only the ratio of the remaining factors determines the result. 

For instance, in the second row, we have $9$ heads and $1$ tails. Given the current bias estimates $\theta_A=0.6$ and $\theta_B=0.5$, the factors are

$$\theta_A^9 (1-\theta_A)^{10-9}\simeq0.004$$

and

$$\theta_B^9 (1-\theta_B)^{10-9}\simeq0.001\;,$$

resulting in the numbers

$$\frac{0.004}{0.004+0.001}=0.8$$

and

$$\frac{0.001}{0.004+0.001}=0.2$$

in the second row.
