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This is from some class notes of a friend. In the line marked as (??), I can’t understand the inclusion of -2. When x belongs to [-2,2], the transformation isn’t 1-1, I get all that. But when the transformation is 1-1, and x belongs to (2,3], why are we putting -2 in the lower limit of the CDF of Y? Shouldn’t it be 2? Or why put a number at all...its all very confusing.

AP _
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2 Answers2

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In the line above, we have $P[-2\leq X\leq \sqrt y]$. Translating this into the CDF of $x$, this is exactly equal to $F_x(\sqrt y) - F_x(-2)$.

To adress the more basic concern: For any $y$-value between $0$ and $4$, the value of $Y$ lies between $0$ and $y$, meaning $X$ lies between $-\sqrt y$ and $\sqrt y$.

When $y\geq 4$, the same bound of being between $-\sqrt y$ and $\sqrt y$ still applies to $X$, but this time we are also bound by the fact that $X\in [-2, 3]$. So $X$ can't go as low as $-\sqrt y$, it can only go as low as $-2$. Any value of $X$ between $-2$ and $\sqrt y$ (and no other values of $X$) is going to give $Y = X^2$ a value between $0$ and $y$, which is what we're after.

Arthur
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  • I understood that portion, I just can’t understand why -2 and not +2? Because when y is varying from 4 to 9, x takes values 2 to 3. The pdf for x ranging from -2 to 2 is already covered separately. So I don’t understand why -2... – AP _ Nov 27 '19 at 14:23
  • I’ve edited the question accordingly – AP _ Nov 27 '19 at 14:30
  • @AP_ I edited my answer a bit. – Arthur Nov 27 '19 at 14:31
  • FINALLY. I’ve been trying to understand this for a few hours, don’t know why it took me so long. Thank you so much. – AP _ Nov 27 '19 at 14:33
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$(*)$ is actually redundant.

It is enough to state that for $y\geq0$ we have $$F_Y(y)=F_X(\sqrt y)-F_X(-\sqrt y)$$ This however in the understanding that $y\geq4$ we have $F_X(-\sqrt y)=0$ and for $y\geq9$ we have $F_X(\sqrt y)=1$.

That comes into play later by taking the derivative in order to find the PDF.

drhab
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