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suppose we have joint probability density function f(x,y)

f(x,y) = 1 if (x,y) is in the triangular area depicted in the sketch.

f(x,y) = 0 elsewhere

Sketch (crudely drawn):

enter image description here

Now we want to convert it to joint CDF. We have a formula for that:

$$\int_{-\infty}^x \int_{-\infty}^y f(u,v) \,dv\,du$$

Now I'm not too sure if I'm applying the lower bounds to this problem correctly: $$\int_{-1}^x \int_0^y f(u,v) \,dv\,du$$

After integrating, I get that in the triangular area, jCDF should be equal to y(x+1).

However, it looks strange because for x=0, y=1, the probability is 1, but should be 0.5 if I'm not mistaken.

Thank you in advance!

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The CDF of $(X,Y)$ is defined as

$$P(X \leq x, Y\leq y)=\int_{A \cap]-\infty,x]*]-\infty,y]}dxdy$$

where $A$ is the surface of the triangle .

In other words, we want to value the surface of $A \cap]-\infty,x]*]-\infty,y]$.

If $(x,y)\in A$, we must evaluate the surface of the trapezoid with edges $(-1,0),(x,0)(x,y)$ and $(y-1,y)$, therefore $$P(X \leq x, Y\leq y)=\frac{(2x-y+2)y}{2}$$

Your result is different, you should double check. When $(x,y)=(0,1)$, the trapezoid is a triangle ( the above formula covers it, the small base is nil), and its surface is $\frac{1}{2}$.

Be careful, you must also look at the cases where $x \notin A$. If $y<0$, the CDF is nil. if $x>1, y>1$, the CDF is equal to one ...

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