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Context of the problem:

Continuous bivariate random variable $(Y_1, Y_2)$ has the uniform density $f(y_1, y_2)$ on support S = $(y_1, y_2) \leq 1-y_1^2, y_1 \leq 0, y_2 \leq 0$. Thus, $f(y_1, y_2)$ has a positive constant value on S and value 0 elsewhere.

The question says "Determine $f(y_1, y_2)$ precisely" but doesn't ask for a specific value. Is this asking for the value of the integral across all of [0,1] where the function is defined?

Nick
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  • What does it mean that the support $S = (y_1,y_2) \leq 1 - y_1^2$? – cgonagu Nov 05 '13 at 10:27
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    It is still unclear what the support is here. If $f$ being uniform means maybe that it is constant. If $f\equiv c$ on $S$ then you need $1=\int_{S}f\left(y_{1},y_{2}\right)dy_{1}dy_{2}=c\lambda\left(S\right)$. Here $\lambda\left(S\right)$ stands for the Lebesgue-measure of $S$. If $S$ (hence $\lambda\left(S\right)$) is known then based on this equality you can determine $c$. – drhab Nov 05 '13 at 10:29

1 Answers1

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If $f\left(y_{1},y_{2}\right)$ is constant on support $S\subset\mathbb{R}^{2}$ and takes constant value $c$ there, then $1=\int\int f\left(y_{1},y_{2}\right)dy_{1}dy_{2}=c\lambda\left(S\right)$ where $\lambda$ denotes the Lebesgue measure on $\mathbb{R}^{2}$.

This way you find $c=\lambda\left(S\right)^{-1}$ and $f$ is determined precisely.

drhab
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