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The probability of heads showing up upon tossing a certain coin is $p$, this coin is tossed $3$ times, let $X_i,i=1,2,3$ be $1$ or $-1$ depending on the outcome of the $i^{th}$ toss being head or tails respectively.Then which of the following statements are true?

$1.X_1+X_2+X_3$ is sufficient statistic for $p$.

$2.X_1^{2}+X_2^{2}+X_3^{2}$ is a sufficient statistic for $p$.

$3.X_1X_2X_3$ is sufficient statistic for $p$.

4.$X_1^{3}+X_2^{3}+X_3^{3}$ is a sufficient statistic for $p$.

I have tried to solve in the following way:

Let $Y_i=\frac{X_i+1}{2}=1$,when $i$th toss is head.

$=0$,when $i$th toss is tail.

Clearly $Y_i$ is a bernoulli random variable and $\sum Y_i$ is sufficient for $p$. But does that imply $\sum X_i$ will also be sufficient?And what about the other statements?

Thanks,in advance.

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    $\begingroup$ If you write down the likelihood you will see which of these is sufficient by checking whether or not the likelihood depends only on these statistics. $\endgroup$ – Xi'an May 29 '16 at 10:02
  • $\begingroup$ @Xi'an, can you please give me some guidance concerning the likelihood function?as distribution of $X_i$ does not come under any known distribution,how to find the likelihood? $\endgroup$ – priyanka May 29 '16 at 10:52
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    $\begingroup$ This is a Bernoulli likelihood, it sounds hard to find a simpler likelihood. $\endgroup$ – Xi'an May 29 '16 at 15:30
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The likelihood of your sample writes as $$\begin{align*} \prod_{i=1}^3 \{p\mathbb{I}_1(x_i)+(1-p) \mathbb{I}_{-1}(x_i)\}&=\prod_{i=1}^3 p^\frac{1+x_i}{2}(1-p)^\frac{1-x_i}{2}\\&=[p(1-p)]^{3/2} p^\frac{\sum_{i=1}^3 x_i}{2}(1-p)^\frac{-\sum_{i=1}^3x_i}{2}\\ &=[p(1-p)]^{3/2} \sqrt{p/(1-p)}^{\sum_{i=1}^3 x_i} \end{align*}$$ From this representation, you can deduce whether or not

  1. $\sum_{i=1}^3 x_i$

    taking four different values and appearing directly in the likelihood: sufficient

  2. $\sum_{i=1}^3 x_i^2$

    a constant equal to 3: insufficient

  3. $\prod_{i=1}^3 x_i$

    only taking two different values, $1$ and $-1$: insufficient

  4. $\sum_{i=1}^3 x_i^3$

    actually equal to $\sum_{i=1}^3 x_i$: sufficient

are sufficient by looking at the values of the likelihood function when those statistics vary.

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    $\begingroup$ This "invisible text" idea is really cool! $\endgroup$ – Zen Jun 4 '16 at 17:59

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