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If X is a discrete uniform random variable with Supp(X) = {−5, −4, . . . , 4, 5} and Y = $X^2$ - 3X, how would you find both the PMF and CDF of Y? I've tried researching how to find PMF and CDF but I still can't seem to understand either and I would really appreciate it if someone could explain it

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First, You need to see that the values of the random variable $Y$ depend on the values of the variable $X$ and it takes integers between $-5$ and $5$. That is,

If $X=-5\Rightarrow Y=(-5)^2-3(-5)=25+15=40$

If $X=-4\Rightarrow Y=(-4)^2-3(-4)=16+12=28$

$\vdots$

If $X=4\Rightarrow Y=(4)^2-3(4)=16-12=4$

If $X=5\Rightarrow Y=(5)^2-3(5)=25+15=10$

Then, $supp(Y)=\{-2,0,4,10,18,28,40\}$.

Second, for the PMF you need to find the probabilities of the variable $Y$ when it takes values in $supp(Y)$, i.e,

$p_Y(y)= \left\{ \begin{array}{lcc} P(Y=40) \\ \\ P(Y=28) \\ \\ \vdots \\ P(Y=-2) \\ \end{array} \right.$

Third, for the CMF you need to find the probabilities of the variable $Y$ when it is less than or equal $y\in supp(Y)$, i.e,

$F(Y)=P(Y\le y)= \left\{ \begin{array}{lcc} P(Y= -2) \\ \\ P(Y\le 0 )= P(Y= -2 )+P(Y= 0 ) \\ \\ \vdots \\ P(Y\le 40)= P(Y= -2 )+P(Y= 0 )+\cdots+P(Y=40 )=1\\ \end{array} \right.$

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  • $\begingroup$ This was really helpful, thanks! But where did you get the -4 from? The lowest number I got for Y was -2 $\endgroup$ – kdd Mar 18 '18 at 23:46
  • $\begingroup$ You're right, I made a mistake when calculating $Y$ values $\endgroup$ – Alex Pozo Mar 19 '18 at 1:28

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