Joint CDF in probability density function Here is the question:
Suppose that the probability density function of random vector $(X,Y)$ is $f(x,y)=\frac{a}{y}$ for $0<x\leqslant y<1$ and $f(x,y)=0$ otherwise. Here $a$ represents an unknown constant.
I have already known the answer of $a$ is $a=1$.
The second question is Find the joint CDF $F(x,y)$, pay attention to the support $0<x\leqslant y<1$.
My answer is:
$$F(x,y)=\int_{0}^{y}\int_{x}^{1}\frac{1}{y}dydx$$
$$=\int_{0}^{y}[ln(y)]_{x}^{1} dx$$
$$=\int_{0}^{y}ln(1)-ln(x) dx$$
$$=\int_{0}^{y}-ln(x)dx$$
$$=[-xln(x)+x+c]_{0}^{y}$$
$$=-yln(y)+y+c$$
I am not sure about my answer to this question. I also have another answer:
$$F(x,y)=\int_{x}^{1}\int_{0}^{y}\frac{1}{y}dxdy$$
$$=\int_{x}^{1}(y/y-0)dy$$
$$=\int_{x}^{1}1 dy$$
$$=1-x$$
I would like to know the correct answer among these two or neither is wrong. Thank you so much!
 A: By definition, the multivariate cumulative distribution function (CDF) is given by
\begin{aligned}
F(x,y) &= \mathbb{P}(X\leq x, Y\leq y) \\
&= \int_{\mathbb{R}^2} \frac{1}{\gamma}\, \mathbf{1}_{0 < \xi \leq \gamma < 1}\, \mathbf{1}_{\xi\leq x}\, \mathbf{1}_{\gamma\leq y}\, \mathrm{d}\xi\, \mathrm{d}\gamma \, .
\end{aligned}
The part of the $\xi$-$\gamma$ plane defined by the inequality $0 < \xi \leq \gamma < 1$ is a triangle, whereas the part defined by the inequalities $\xi \leq x$ and $\gamma \leq y$ is a quarter-plane. Both parts intersect if $x>0$ and $y>0$ (which is assumed from now on), otherwise the integral is zero. The computation of the integral follows from Fubini's theorem:
\begin{aligned}
F(x,y) &= \int_{\mathbb{R}} \frac{1}{\gamma}\, \mathbf{1}_{\gamma\leq y}\, \mathbf{1}_{\gamma < 1} \left( \int_{\mathbb{R}} \mathbf{1}_{0 < \xi \leq \gamma}\, \mathbf{1}_{\xi\leq x}\, \mathrm{d}\xi\right) \mathrm{d}\gamma \, ,\\
&= \int_{\mathbb{R}} \frac{1}{\gamma}\, \mathbf{1}_{\gamma\leq y}\, \mathbf{1}_{0 <\gamma < 1} \left( \int_{0}^{\min\lbrace x,\gamma\rbrace} \mathrm{d}\xi\right) \mathrm{d}\gamma \, , \\
&= \int_{0}^{\min\lbrace y,1\rbrace} \frac{\min\lbrace x,\gamma\rbrace}{\gamma} \, \mathrm{d}\gamma \, .
\end{aligned}
Several cases must be considered:


*

*if $x\geq \min\lbrace y,1\rbrace$, then $$ F(x,y) = \int_0^{\min\lbrace y,1\rbrace} \mathrm{d}\gamma = \min\lbrace y,1\rbrace \, . $$

*else,
$$
F(x,y) = \int_0^x \mathrm{d}\gamma + x\int_x^{\min\lbrace y,1\rbrace} \frac{\mathrm{d}\gamma}{\gamma} = x + x \ln\left(\frac{\min\lbrace y,1\rbrace}{x}\right) .
$$

