Joint cumulative distribution function

Linder88
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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}
 
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Why are these variables not independent?

If they aren't, then Fxy is not equal to FxFy
 
My teacher told they are not independent even though I wish they were :frown:
 
Linder88 said:
My teacher told they are not independent even though I wish they were :frown:

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

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Linder88 said:
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##.
 
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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}
 
Linder88 said:
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?
 
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  • #10
Ray Vickson said:
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.
 
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