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Given a discrete stochastic variables, with the probability function; $$ p_{X}\left(x\right)=\left\{ \begin{array}{cc} \frac{1}{4} & \text{if } x = -1 \\ \frac{1}{4} & \text{if } x = 0 \\ \frac{1}{2} & \text{if } x = 1 \\ 0 & \text{otherwise} \end{array} \right. $$ How do I prove that; $$|X| \sim b\left(1, \frac{3}{4}\right)$$ What I've done so far: Found $P(|X| = 1) = \frac{3}{4}$, and then I've unsuccessfully tried to derive the probability function (above) from the binomial probability function (below); $$ p\left(x\right) = \left( \begin{array}{c} n \\ x \end{array}\right) \pi^{x} \left(1 - \pi\right)^{n-x} $$ However without any luck, as $|X| \neq X$.

So my question is somewhat simple; How would I go about proving that a discrete stochastic variable is binomial distributed?

Skeen
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1 Answers1

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You already did all of the work. You showed that $$ p_{\left|\mathbf{X}\right|}\left(x\right) = \left(\begin{array}{c} 1\\ x \end{array}\right)\frac{3}{4}^{x}\frac{1}{4}^{1-x} = \begin{cases} \frac{3}{4} & \text{if }x=1\\ \frac{1}{4} & \text{if }x=0 \end{cases} $$

parsiad
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  • Wouldn't that be $ p_{|X|}\left(x\right) = \left( \begin{array}{c} 1 \ x \end{array}\right) \frac{3}{4}^{x} \frac{1}{4}^{1-x} $? - Also why is the support = {0,1}? - Is it the support is the same as $p_{X}\left(X\right)$, however with $-1$ excluded due to it's undefinedness in factorial? – Skeen Aug 08 '13 at 20:48
  • Also, how does this prove that $|X|$ is binomial distributed? - Because it is possible to write it to binomial distributed form, while retaining the $P(|X| = 1)$? – Skeen Aug 08 '13 at 20:54
  • Yes to the correction. As for the support, $p_{\left|\mathbf{X}\right|}\left(x\right)=0$ for anything other than $x=0$ or $x=1$. We say an r.v. is binomially distributed if its probability mass function at $k$ is given by $\left(\begin{array}{c}n\k\end{array}\right)p^{k}\left(1-p\right)^{n-k}$ for some $p$ and $n$. You showed this (specifically, with $n=1$ and $p=3/4$). – parsiad Aug 08 '13 at 21:03
  • I'm still uncertain about why the support is $= {0,1}$? – Skeen Aug 08 '13 at 22:27
  • The support of a function is defined as the set of points on which the function is nonzero. The function is nonzero only at $x=1$ and $x=0$. – parsiad Aug 08 '13 at 23:27
  • I get that, however, I'm uncertain about how to find these values? That is how do I determine which $x$ to test for non-zero probability? - Trial and error? - Intuition? – Skeen Aug 09 '13 at 11:18