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.
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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.
drhab
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if y=u−x how is dy = dx (because x and y are not independent) – user263042 Aug 19 '15 at 11:19
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1$y=u-x$ leads to $dy=du$ (so not $dy=dx$). Secondly dependent or not dependent is irrelevant here. The $x,y$ are not random in this setting, but are just variables ranging over $\mathbb R$. We are dealing with an integral that is manipulated. – drhab Aug 19 '15 at 11:28
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I guess it to be dy = du - dx – user263042 Aug 19 '15 at 12:51
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1No. $dy=du$. In equation $y=u-x$ the $x$ is treated as a constant. Then $\frac{dy}{du}=\frac{d(u-x)}{du}=1$ or equivalently $dy=du$. – drhab Aug 19 '15 at 13:42
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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:
$$f_{W}(w)=\int_{0}^{\infty}f_{W,Z}(w,z)dz$$
$$f_{Z}(z)=\int_{0}^{\infty}f_{W,Z}(w,z)dw$$
Note:$a,b,c,d$ are defined in the way that: $$ad-bc\neq 0$$
Cardinal
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yes, your question is very simple version of my answer ! I mean your answer can be easily deduced from my answer. you need to set A=B=1. – Cardinal Aug 19 '15 at 11:23
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I didn't understand what your saying because my question has only three random variables X,Y and Z=X+Y . What do W mean? – user263042 Aug 19 '15 at 11:26
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@user263042 $W$ is a temporary random variable that I considered to use Jacobian method – Cardinal Aug 19 '15 at 12:58
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Let $X,Y$ be dependant variables with PDF $f(x,y)$.
Then sum PDF is: $$f_{X+Y}(k)= \int f(x,k-x) dx$$
hvedrung
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If $f(x,y)=P(X=x,Y=y)$, you probably mean $P(X+Y=k)=\sum\limits_xP(X=x,Y=k-x)=\sum\limits_xf(x,k-x)$, not $P(X+Y=k)=\int f(x,k-x)dx$. – Did Aug 19 '15 at 11:18
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