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Suppose $x$ and $y$ are uniformly distiruted between $[-L;L]$, what is the probability that $x<y^2$?

I found this 5-year old discussion: http://answers.yahoo.com/question/index?qid=20080308204357AAEdyRP&r=w but it is long and lot very illustrative.

Edit: OK, I understand the idea. I just don't understand why there is need to seperate $L<1$?

Why can't you just take and integrate $y^2$ in interval $[-L,L]$ then add $2L^2$ (area below 0) and divide everything with $4L^2$ (all area)?

Karlis Olte
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  • answer to the edit : See my drawing : the parabola intersects the square differently when L>1. – justt Jun 11 '13 at 22:28

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If $L\le 1$, we find

$$\frac1{(2L)^2}\int_{-L}^L (L+y^2)\,\mathrm dy =\frac1{2}+\frac16L.$$

If $L\ge 1$, we find $$\frac1{(2L)^2}\left(\int_{-L}^{-\sqrt L}2L\,\mathrm dy+\int_{-\sqrt L}^{\sqrt L} (L+y^2)\,\mathrm dy +\int_{\sqrt L}^{L}2L\,\mathrm dy\right)=1 -\frac1{3\sqrt L}.$$

Hagen von Eitzen
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    For $L=4$ this gives $160/3$, which is massively larger than $1$. Whatever you calculated, it's not a probability. I guess you should divide by $4L^2$. Also, your first integral result should have $2L^2$ instead of $2L$. – celtschk Jun 11 '13 at 22:08
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    @celtschk Oops, thanks for pointing that out. I just calculated the area (cf. justt's drawing), not its proportion – Hagen von Eitzen Jun 11 '13 at 22:17
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To answer the question about why to separate $L<1$ from $L>1$, here's how it looks like for $L=1$ ($x$ is the horizontal axis here):

L=1

This is how it looks for $L=1/2$:

L=1/2

And this is how it looks for $L=2$.

L=2

As you can see, for $L\le1$, the whole right edge is cut away, while for $L>1$ part of the edge remains.

celtschk
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illustration case L>1

Here I tried to illustrate the case $L>1$.

justt
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