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This was an example problem my professor went over in class.

Let $X =$ uniform $(1,4)$ where $Y=(X-2)^2$ Find the CDF.

He went on to derive:

$F_Y(y)=P(Y\leq y) = P((x-2)^2 \leq y) = P(-\sqrt{y}\leq (x-2) \leq \sqrt(y))$

= $P(2 - \sqrt{y} \leq x \leq 2 + \sqrt{y})$

Then he said the CDF is:

$$F_X(x)= \begin{cases} 0 & x\leq 1 \\ \frac{x-1}{3} & 1\leq x\leq 4 \\ 1 & x\geq 4 \end{cases} $$

I am very confused as to how he derived this CDF. Could someone please explain?

Edit: Unless he is wrong?

Mark
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  • CDF for $F_X(x)$ is simply the area of the rectangle of height $\frac{1}{3}$ and basis from (1,0) to (x,0) (lenght $x-1$) – Jonas Gomes Dec 05 '19 at 04:47

1 Answers1

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The CDF given was simply that of the variable $X$, it's not the CDF for $Y$. Note that since this is a uniform distribution on $(1,4)$ then the density is given by $p_X(x) = \frac{1}{3}$. Then:

$$ F_X(x) \;\; =\;\; P(X\leq x) \;\; =\;\; \int_1^x \frac{1}{3}d\alpha \;\; =\;\; \frac{x-1}{3}. $$

The values on the other parts of the domain are because $P(X\leq 1) = 0$ and $P(X\geq 4) = 1$.

Mnifldz
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  • @Mnifidz Thank you, but why did you do limits from 1 to x? – Mark Dec 05 '19 at 04:56
  • @Mark See my edits above. It's because the CDF is defined by $P(X\leq x)$. – Mnifldz Dec 05 '19 at 04:58
  • @Mnifidz Ahh im sure what you're saying makes sense but I am still really confused as to how you got those bounds. – Mark Dec 05 '19 at 05:08
  • I cant think logically how you got them – Mark Dec 05 '19 at 05:09
  • @Mark It follows from how CDFs are defined. Since the distribution is supposed to be on $(1,4)$, then if you try to compute $P(X\leq a)$ where $a\leq 1$, then it falls outside the interval $(1,4)$ hence the probability should be zero that you sample a value in that range. On the other hand, if you compute $P(X\leq a)$ where $a\geq 4$, it covers the whole sample space, hence there's probability $1$ that you obtain a sample in that range. – Mnifldz Dec 05 '19 at 05:12
  • Thank you, so why did the professor even both doing all of the interval manipulation in the first part of the answer? – Mark Dec 06 '19 at 14:35
  • @Mark How should I know? Ask him/her. – Mnifldz Dec 06 '19 at 14:47