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I don´t really know understand this question and as such don´t know what to do.

Let X be a random discrete variable with the probability function: $p_X(x)=\tfrac 18[4~\delta_{(x=-1)}+2~\delta_{(x=0)}+\delta_{(x\in[1,2])}]$ where $\delta$ is the Kronecker delta.

Find the probability function of $Y=2X+1$ and $Y=2X^2 +1$ Thanks in regard.

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    PS: if $X$ is discrete, should not that last term be $\delta_{(x\in\{1,2\})}$ ? – 2017-01-29

1 Answers 1

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Given the probability mass function $p_X (x)=\tfrac 18 [4 \delta_{(x=-1)}+2\delta_{(x=0)}+\delta_{(x\in\{1,2\})}]$

When we have $Y=2X+1$ we have a bijection, $\{-1,0,1,2\}\mapsto\hspace{-2ex}\to \{-1,1,3,5\}$, so we simply substitute

$$p_{(2X+1)}(y) = \tfrac 18 [4 \delta_{(y=-1)}+2\delta_{(y=1)}+\delta_{(y\in[3,5])}]$$

Although we don't have a bijection for $Y=2X^2+1$, where $\{\color{blue}{-1},0,\color{blue}{1},2\}\mapsto \{\color{blue}3,1,\color{blue}3,9\}$ we substitute and collate like terms

$$p_{(2X^2+1)}(y) = \tfrac 18 [5 \delta_{(y=3)}+2\delta_{(y=1)}+\delta_{(y=9)}]$$