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Let $X$ ~ Unif$([0,10])$ be a continuous random variable. I now want to find the probability mass function of $Y :=- \frac{1}{2}X+4$. I did some reading and figured out that $Y$ ~ Unif$([-1,4])$ with $f_{Y}(y)=f_{X}(-2y+8)\cdot 2 \cdot 1_{[-1,4]}$. So far, so good.

However, I have no idea how to find the probability mass function and distribution function of $\sqrt{X}$ and $X^{2}$. I would be grateful for any tip into the right direction. Does it work the same way as with linear transformations of random variables?

Jolle
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2 Answers2

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Hint:

Find CDF's of $\sqrt X$ and $X^2$.

Then find PDF's (not PMF as you call it) by differentiating the CDF's.

drhab
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  • Thank you! I know that the CDF of $X$ is given by $F_{X}(u)=\frac{u}{10}$ for $0 \le u \le 10$. Do I have to work with this to find the CDFs of $X^{2}$ and $\sqrt{X}$? – Jolle Dec 27 '18 at 12:59
  • @Jolle Writing down the expression for the CDF of $X^2$ and $\sqrt X$ would make that clear. – StubbornAtom Dec 27 '18 at 13:17
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    As @StubbornAtom says. For nonnegative $x$ just start with $F_{\sqrt X}(x)=P(\sqrt X\leq x)=P(X\leq x^2)=\cdots$ – drhab Dec 27 '18 at 13:28
  • Thank you! I continue: – Jolle Dec 28 '18 at 11:42
  • Thank you! I continue: $... = P(X \le x^{2}) = F_{X}(x^{2})=\frac{x^{2}}{10}$ for $0 \le x^{2} \le 10$ right? Now I differentiate: $f_{\sqrt{X}}(x)=(F_{X}(x^{2}))‘=\frac{1}{5}x$. Is this correct? – Jolle Dec 28 '18 at 11:49
  • Yes for $x\in[0,\sqrt{10}]$. Outside that interval let it take value $0$. – drhab Dec 28 '18 at 11:52
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Transformation works the same way, linear or not.

If $Y=g(X)$, then $x=g^{-1}(y)$ if unique inverse exists (as in both these cases). So you have a one-to-one transformation from $X$ to $Y$. (This would not be the case in general)

Find the support $S^*$ of $Y$ from the support of $X$.

Then by change of variables, probability density function of $Y$ is

$$f_Y(y)=f_X(g^{-1}(y))\left|\frac{d}{dy}g^{-1}(y)\right|\mathbf1_{y\in S^*}$$

For linear transformations, the jacobian $\frac{d}{dy}g^{-1}(y)$ is constant. Here that is not the case.

StubbornAtom
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