Let $X_{1},X_{2},\ldots,X_{n}$ be random sample from a population whose distribution is given by $X\sim\text{Bernoulli}(\theta)$, $0 < \theta < 1$.

a. Show that $T(x) = \displaystyle\sum_{i=1}^{n}X_{i}$ is sufficient for $\theta$.

b. In the same context, consider that $T = X_{1} + X_{2}$. What is the distribution of $T$? Is $T$ a sufficient statistic?


a. To begin with, let us determine the likelihood function for this sample \begin{align*} L(\textbf{x}|\theta) = \prod_{i=1}^{n}\theta^{x_{i}}(1-\theta)^{1-x_{i}} = \theta^{\sum x_{i}}(1-\theta)^{n - \sum x_{i}} = h(\textbf{x})g_{\theta}(T(\textbf{x})) \end{align*}

Therefore, according to the factorization theorem, $T(\textbf{x})$ is sufficient for $\theta$.

b. As it is known, the sum of independent Bernoulli random variables is a Binomial random variable. Therefore $T = X_{1} + X_{2} \sim \text{Binomial}(2,\theta)$.

Then I get stuck. Can someone help me out? Thanks in advance!

  • $\begingroup$ (a) trivially implies the answer to the second part of (b) by setting $n=2.$ $\endgroup$ – whuber May 17 '19 at 15:12
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    $\begingroup$ On the other hand, if $n>2$ then the answer by @Vishaal answer is helpful. I think this may be the intention of the question, but it is unclear with the information you have given us. $\endgroup$ – knrumsey May 17 '19 at 17:34

All you need to show is that $$P(X_1=x_1,X_2=x_2,...X_n=x_n|X_1+X_2 = t )$$ depends on $\theta$ to prove that $T=X_1+X_2$ is NOT a sufficient statistic.

  • $\begingroup$ Could you explain what part of the question this answers and why it answers it? $\endgroup$ – whuber May 17 '19 at 15:13
  • $\begingroup$ @Whuber I have answered "Is $X_1 + X_2$ sufficient?" ... That was the last question where the OP got stuck. I interpreted the question like this, out of a random sample following bernoulli($\theta$)of size N if we know the sum of the first 2 random variables does it contain sufficient information about the parameter $\theta$. $\endgroup$ – Vishaal Sudarsan May 17 '19 at 15:34
  • $\begingroup$ Unfortunately, that answer is wrong. It contradicts the answer to question (1). $\endgroup$ – whuber May 17 '19 at 16:07
  • $\begingroup$ According to me.. Question(1) is whether sum of all samples is sufficient of not. Question(2) is, In the presence of N samples, whether the sum of first 2 sufficient or not. I've updated my answer. Please check it once more. $\endgroup$ – Vishaal Sudarsan May 17 '19 at 16:24
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    $\begingroup$ In that case you're correct. But I understood the question as we have already observed n samples, now before discarding the sample we would like to compute a sufficient statistic for $\theta$, so is $X_1 +X_2$ a sufficient statistic?. I interpreted the question in this way because in Question(2) it says "In the same context" so I assumed the OP wants to say all n samples have been observed like it has been observed for Question(1). $\endgroup$ – Vishaal Sudarsan May 17 '19 at 17:13

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