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Suppose that $X$ and $Y$ are independent and have beta distributions. $X$ has probability density function $g(x)=6x(1-x)$ for $ 0\leq x \leq 1$ and $Y$ has the probability density function $h(y)=12y^2(1-y)$ for $0\leq y \leq 1$. I want to find $P(X+Y\leq 1)$.


I have calculated $f(x,y)=72x^2(1-x)y^2(1-y);0 \leq y \leq 1$. How do I compute $p(X+Y\leq 1)$? Expert's guidance is needed. The answer given to me is $\frac{13}{35}$, but how is it computed?

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This question is squarely about statistics and conforms to the self-study guidelines. I vote to keep it open. – General Abrial Mar 3 at 16:57

Not an expert but here are some hints:

1) You do not actually need to derive the pdf of the $X+Y$ but merely perform the integration over the set $\left\{ \left(x,y\right): x+y \leq 1 , \ 0 < x,y < 1 \right\}$, which you can plot to get the idea.

2) The area of integration turns out to be quite convenient as you can use the result

$$\int_{0}^1 x^{a-1} \left(1-x\right)^{b-1} \mathrm{dx} = \frac{\Gamma(a) \Gamma(b) }{ \Gamma (a + b) }$$

where $\Gamma(x)$ is the gamma function with the property that $\Gamma(x) = (x- 1)!$. You should be able to get to the result now.

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Then answer is easy and comes from the fact that $p(a,b)=p(a)p(b|a)$ then

$$ P(X+Y\leq 1) = \int_x P(X=x)P(Y\leq 1-x) $$ if you replace $P(X)$ and $P(Y)$ by $g(x)$ and $h(y)$ then you get the result. $$ P(X+Y\leq 1) = \int_0^1 g(x)\int_0^{1-x}h(y)dydx $$ The final answer would be $72*(1/90-1/168)$.

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Well, strictly speaking $P(X=x) = 0$... – JohnK Mar 3 at 23:52
I see but the underlying assumption is small intervals for continues variables. (we usually do not mention that) – TPArrow Mar 12 at 15:42

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