pdf of sum of two dependent random variables

Question

What is the pdf of sum of two dependent random variables given we know their joint pdf and individual pdfs. I have seen already some posts but none of them answered when they are dependent. Every one solved for only the independent case but i need for dependent case in terms of the joint pdf and individual pdfs in an explicit form.

Answer 1

If $f(x,y)$ denotes the PDF of $(X,Y)$ and $Z=X+Y$ then: $$f_Z(z)=\int^{\infty}_{-\infty}f(x,z-x)dx$$

Substituting $y=u-x$ we find:$$F_Z(z)=\int\int_{x+y\leq z}f(x,y)dxdy=\int^{\infty}_{-\infty}\left[\int^{z-x}_{-\infty}f(x,y)dy\right]dx=$$$$\int^{\infty}_{-\infty}\left[\int^{z}_{-\infty}f(x,u-x)du\right]dx=\int^{z}_{-\infty}\left[\int^{\infty}_{-\infty}f(x,u-x)dx\right]du$$

Taking the derivative of $F_Z$ we come to the mentioned result.

Answer 2

You can use Jacobian determinant and temporary variables. Consider :

$$Z=aX+bY$$

$$W = cX+dY$$

Therefore, we get:

$$X=AZ+BW\ \ \ \ \ , \ \ \ \ Y=CZ+DW$$

And for a given pair of $X=x$ and $Y=y$, we have one solution as:

$$x=Az+Bw\ \ \ \ \ , \ \ \ \ y=Cz+Dw$$

Thus Jacobian determinat would be:

$$J(x,y)=ad-bc$$

which translates to:

$$f_{W,Z}(w,z)=\frac{1}{|ad-bc|} f_{x,y}(Az+Bw, Cz+Dw)$$

Here after, you can calculate the marginal PDF of W or Z by a simple integration. In other words:

$$f_{W}(w)=\int_{0}^{\infty}f_{W,Z}(w,z)dz$$

$$f_{Z}(z)=\int_{0}^{\infty}f_{W,Z}(w,z)dw$$

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Note:$a,b,c,d$ are defined in the way that: $$ad-bc\neq 0$$

Answer 3

Let $X,Y$ be dependant variables with PDF $f(x,y)$.

Then sum PDF is: $$f_{X+Y}(k)= \int f(x,k-x) dx$$