# Explanation of “sufficient statistic” example

I just started learning what a sufficient statistic is:

Definition

A statistic $$T(\mathbf{Y})$$ is sufficient for an unknown parameter $$\theta$$ if the conditional distribution of the data $$\mathbf{Y}$$ given $$T(\mathbf{Y})$$ does not depend on $$\theta$$.

Example

Let $$Y_1, \dots, Y_n$$ be a i.i.d. $$B(1, p)$$. Let $$T(\mathbf{Y}) = \sum_{i = 1}^n Y_i$$ be a statistic. Show that $$T(\mathbf{Y})$$ is a sufficient statistic. Note that $$T(\mathbf{Y}) \sim B(n, p)$$ and therefore,

\begin{align} P(\mathbf{Y} = \mathbf{y} \vert T(\mathbf{Y}) = t) &= P(\mathbf{Y} = \mathbf{y} \vert \sum_{i = 1}^n Y_i = t) \\ &= \dfrac{P(Y_1 = y_1, \dots, Y_n = y_n, \sum_{i = 1}^n Y_i = t)}{P(\sum_{i = 1}^n Y_i = t)} \\ &= \dfrac{P(Y_1 = y_1, \dots, Y_n = y_n)}{P(\sum_{i = 1}^n Y_i = t)} \\ &= \dfrac{p^{\sum_{i = 1}^n y_i}(1 - p)^{n - \sum_{i = 1}^n y_i}}{{n\choose{t}}p^{\sum_{i = 1}^n y_i}(1 - p)^{n - \sum_{i = 1}^n y_i}} = \dfrac{1}{{n\choose{t}}} \end{align}

that does not depend on the unknown $$p$$.

I'm having difficulty following this example, and I would appreciate it if people could please help me understand what's going on here.

1. How is it that $$P(Y_1 = y_1, \dots, Y_n = y_n, \sum_{i = 1}^n Y_i = t) = P(Y_1 = y_1, \dots, Y_n = y_n)$$ in the numerator? What happened to $$\sum_{i = 1}^n Y_i = t$$?

2. What is the reasoning the led from $$\dfrac{P(Y_1 = y_1, \dots, Y_n = y_n)}{P(\sum_{i = 1}^n Y_i = t)}$$ to $$\dfrac{p^{\sum_{i = 1}^n y_i}(1 - p)^{n - \sum_{i = 1}^n y_i}}{{n\choose{t}}p^{\sum_{i = 1}^n y_i}(1 - p)^{n - \sum_{i = 1}^n y_i}}$$? I understand that the binomial probability mass function was used here for the joint probability, but, since I'm not experienced with joint probabilities, I'm still not clear on the reasoning involved here. Furthermore, I don't understand why the binomial coefficient $${n\choose{t}}$$, which is part of the binomial probability mass function, was used in the denominator but not the numerator.

Thank you.

For 1. the expression is for joint probability. If you already know the specific values of each of the $$Y_1, \ldots, Y_n$$, then their sum is trivial information.
For 2. the probability distribution for the sequence of Bernoulli RVs is just the product of $$n$$ times the Bernoulli PDF since they're IID. Remember, in the numerator we know $$Y_1$$'s specific value whereas for the Bernoulli, we just know how many $$Y$$'s were 1, but $$Y_1$$ may or may not actually be one. The sum of Bernoulli RVs has a binomial distribution function. All that remains is the combinatorial mishmash that clarifies the actual sequence of 0s and 1s... and that has nothing to do with $$p$$.
• For 2., I misread that the $Y_i$ are i.i.d Bernoulli random variables. And since they are i.i.d., this en.wikipedia.org/wiki/… property means that the joint distribution is equal to the product of the random variables; and the numerator $p^{\sum_{i = 1}^n y_i}(1 - p)^{n - \sum_{i = 1}^n y_i}$ is the product of those $n$ Bernoulli random variables (that is, the product of their probability mass functions). For the denominator ${n\choose{t}}p^{\sum_{i = 1}^n y_i}(1 - p)^{n - \sum_{i = 1}^n y_i}$, [...] – Dom Fomello Mar 5 '20 at 3:21
• [...] we must remember that the sum of $n$ i.i.d. Bernoulli random variables, which is what $P(\sum_{i = 1}^n Y_i = t)$ is, is equal to a binomial random variable. So we get the result that the denominator ${n\choose{t}}p^{\sum_{i = 1}^n y_i}(1 - p)^{n - \sum_{i = 1}^n y_i}$ is the binomial probability mass function, where the probability of success is the sum of all the $y_i$. – Dom Fomello Mar 5 '20 at 3:28