A discrete random variable $Y$ has probability function $p(y) = 1/5$ for $y = -2, -1, 0, 1, 2$. Consider the random variable $U = Y^2$. What is the probability function (call it $p_*(u)$) for $U$? The only problem I'm having is tabulating the probability for $p(4)$, any ideas?
Find probability function given a random variable.
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1 Answers
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I think it´s simply: $p_*(0)=1/5,p_*(1)=2/5,p_*(4)=2/5$ and $p_*(k)=0, \forall k\notin\{0,1,4\}$. You can see it like it's usually seen in probability: $p_*(i)=\mathbb{P}(U=i)=\mathbb{P}(Y^2=i)$. Then, e.g. with $i=4$ we have that $p_*(4)=\mathbb{P}(Y^2=4)$ and because of $Y^2=4$ iff $Y=2 \vee Y=-2$, disjoints events, then $\mathbb{P}(Y^2=4)=\mathbb{P}(Y=2)+\mathbb{P}(Y=-2)=1/5+1/5=2/5$