So I got stuck at the following problem: Theres a bivariate random variable with joint pdf $$f_{X,Y}(x,y)=\frac{e^\frac{-(x^2+y^2)}{2\sigma}}{2\pi\sigma^2}$$ $$(-\infty\lt x,y\lt \infty)$$ I was asked to find $P(X,Y)$ while given$\quad$ $x^2+y^2\le a^2$. please help me ! How do I set up the boundaries for my integral when calculating the CDF???
conditional probability with joint pdf
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$\begingroup$
probability
random-variables
density-function
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0$x^2+y^2\le a^2$ implies $y^2\leq a^2-x^2 \iff x\in[-\sqrt{(a^2-y^2),}\sqrt{(a^2-y^2)}]$, you can use these boundaries. Or, in case you are familiar with polar coordinates, you can also use these by doing a coordinates transformation, as the given inequality states the pair $(x,y)$ should be within a circle with radius $a$. – 2017-01-27
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0I am not very clear, do you need the PDF or the CDF? If it is the CDF, do you need it for $x^2+y^2\le a$ or what you need is the CDF of $(X,Y)|X^2+Y^2\le a$? The last one seems to be very messy, with a lot of cases, – 2017-01-27
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0What i need is to find the CDF of (X,Y)|X^2+Y^2
– 2017-01-27
1 Answers
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It is not clear what you mean. Given pdf $f_{X,Y}$, pdf conditional on $x^2+y^2\leq a^2$ is $f_{X,Y|X^2+Y^2\leq a^2}=\frac{f_{X,Y}}{\mathbb{P}[X^2+Y^2\leq a^2]}$.
You mean how do you calculate $\mathbb{P}[X^2+Y^2\leq a^2]$? Just observe $x^2+y^2\leq a^2$ is a circle with center at $(0,0)$ and radius $a^2$. So you take $x\in[-a,a]$ and for each $x$ you take $y\in[-\sqrt{(a^2-x^2)},\sqrt{(a^2-x^2)}]$.