Consider the random variables $Y$ characterised by a probability mass function (PMF) as follows: $$ Y=\begin{cases} -2 & \text{ with probability $\frac{1}{2}$}\\ 2 & \text{ with probability $\frac{1}{2}$}\\ 10 & \text{ with probability $0$} \end{cases} $$ Is this PMF considered symmetric around zero? I'm confused because if I picture it it looks symmetric around zero (the last mass point does not appear as it has probability mass zero). However, from reading the definition of symmetric probability mass function, it does not seem so.
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A result with probability zero is not in the sample space and can thus be discarded no? – Philip Mar 05 '19 at 12:34
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1@Philip, that's a quite bold argument. Think about continuous distributions. – gunes Mar 05 '19 at 12:39
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Sorry my mistake read this like a slot machine with 3 payouts. Thanks for the correction. – Philip Mar 05 '19 at 12:43
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It is symmetric. One could also add $Y=5$ with probability $0$ to the definition as an option, too. Both formal and informal definitions in your link support this. The formal definition is finding a $x_0$ such that $$f(x_0+\delta)=f(x_0-\delta) \ \ \forall \delta$$ This $x_0$ here is $0.$
gunes
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Thanks, but what happens for $\delta=10$? How is it f(10)=f(-10) if $-10$ is not considered in my definition of $Y$? Are we implicitly augmenting the support of $Y$ including $-10$? I think there is some underlying argument on the definition of support that I'm missing. I want to focus on discrete random variables. – TEX Mar 05 '19 at 12:43
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1Normally, your definition should have $-2$ and $2$, and an else statement, stating that we have zero probability when our variable is none of the above. For discrete RVs, PMF is actually a probability: $f_X(x)=P(X=x)$. So, whether you define it or not, $P(Y=-10)$ is $0$. – gunes Mar 05 '19 at 12:46