Joint cumulative distribution function

[Linder88]

Homework Statement

Compute the joint cumulative distribution function $F_XY(x,y)$?

Homework Equations

The marginal distribution function $F_X(x)$
\begin{equation}
F_X(x)=P(X\leq x)=
\begin{cases}
0,x<0\\
0.6,0\leq x<1\\
1,x\geq 1
\end{cases}
\end{equation}
and $F_Y(y)$
\begin{equation}
F_Y=
\begin{cases}
0,y<0\\
0.3,0\leq y<1\\
0.7,1\leq y <2\\
1,y\geq 2
\end{cases}
\end{equation}

The Attempt at a Solution

For independent (I know the are not) random variables X and Y
\begin{equation}
F_XY(x,y)=F_X(x)F_Y(y)=\\
[0.6u(x)+0.4u(x-1)][0.3u(y)+0.4u(y-1)+0.3u(y-2)]=\\
0.6*0.3u(x)u(y)+0.6*0.4u(x)u(y-1)+0.6*0.3u(x)u(y-2)+0.4*0.3u(x-1)u(y)+0.4*0.4u(x-1)u(y-1)0.4*0.3u(x-1)u(y-2)=\\
0.18u(x)u(y)+0.24u(x)u(y-1)+0.18u(x)u(y-2)+0.12u(x-1)u(y)+0.16u(x-1)u(y-1)+0.12u(x-1)u(y-2)
\end{equation}

 

[ElijahRockers]

Why are these variables not independent?

If they aren't, then F_xy is not equal to F_xF_y

 

[Linder88]

My teacher told they are not independent even though I wish they were

 

[Ray Vickson]

My teacher told they are not independent even though I wish they were

If all you are told are the two marginals, then it is impossible to give the joint distribution. Are you not told anything else at all about the two random variables?

 

[WWGD]

I think you need a copula https://en.wikipedia.org/wiki/Copula_(probability_theory) or something related.

 

[Linder88]

Well, the whole question reads like in the attached picture but I already did the first part!

 

[Ray Vickson]

Well, the whole question reads like in the attached picture but I already did the first part!

Just apply the DEFINITION of the joint cdf ##F_{XY}(x,y)##. You will be able to present the results in a ##2 \times 3## table of ##F(x,y)## values, corresponding to ##x = 0,1## and ##y = 0,1,2##.

 

[Linder88]

I guess you mean
\begin{equation}
F_{XY}(x,y)=
\begin{cases}
(0.2+0.3+0.1)(0.2+0.1),x=0;y=0\\
(0.2+0.1+0.1)(0.3+0.1),x=1;y=1\\
0.2+0.1,y=2
\end{cases}
\end{equation}

 

[Ray Vickson]

I guess you mean
\begin{equation}
F_{XY}(x,y)=
\begin{cases}
(0.2+0.3+0.1)(0.2+0.1),x=0;y=0\\
(0.2+0.1+0.1)(0.3+0.1),x=1;y=1\\
0.2+0.1,y=2
\end{cases}
\end{equation}

No, I do not mean that. For one thing, it is completely wrong.

Let me repeat my previous question: what is the DEFINITION of ##F_{XY}(x,y)##?

Expanded question: for a given pair ##(x,y)##, how would you compute that?

 

[Linder88]

No, I do not mean that. For one thing, it is completely wrong.

Let me repeat my previous question: what is the DEFINITION of ##F_{XY}(x,y)##?

Expanded question: for a given pair ##(x,y)##, how would you compute that?

I think i finally get it. For a given pair i would have that
$$
F_{XY}(x,y)=
\begin{cases}
0,x<0,y<0\\
0.2+0.1,0\leq x<1,0\leq y<1\\
0.2+0.1+0.3,0\leq x<1,1\leq y<2\\
0.2+0.1+0.3+0.1,1\leq x<2,1\leq y<2\\
0.2+0.1+0.3+0.1+0.2+0.1,1\leq x,2\leq y
\end{cases}
$$
or
$$
F_{XY}(x,y)=
\begin{cases}
0,x<0,y<0\\
0.3,0\leq x<1,0\leq y<1\\
0.6,0\leq x<1,1\leq y<2\\
0.7,1\leq x<2,1\leq y<2\\
1,1\leq x,2\leq y
\end{cases}
$$
Thanks.