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When the variables $X, Y$ are independent, then the PDF of $Z = X + Y$ can be computed using convolutions: $$ f_Z(z) = \int_{-\infty}^{\infty} f_X(x)f_Y(z - x) dx $$

When the variables are dependent, apparently you can use $$ f_Z(z) = \int_{-\infty}^{\infty} f_{XY}(x, z - x) dx $$

I am wondering where the expression came from for the dependent case? It looks very similar to the independent case except you can't separate the joint distribution into marginals.

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  • $\begingroup$ The derivation/source for both equations is exactly the same, except for independent r.v.'s you can factor the density as a last step. $\endgroup$ Commented May 13, 2021 at 21:43
  • $\begingroup$ @AaronHendrickson Hmm based on Snoop's answer belong, I don't see how it's the same. I can derive the independent case purely from conditional probability: $$ F_{Z}(z) = P(X + Y \leq z) = \int_{-\infty}^{\infty} P(X + Y \leq z | X = x) f_X(x) dx \\ = \int_{-\infty}^{\infty} P(x + Y \leq z) f_X(x) dx \\ = \int_{-\infty}^{\infty} P(x + Y \leq z) f_X(x) dx \\ = \int_{-\infty}^{\infty} P(Y \leq z - x) f_X(x) dx \\ = \int_{-\infty}^{\infty} \int_{-\infty}^{z - x} f_Y(y) f_X(x) dydx \\ $$ Then I think we just take the derivative with respect to $z$ and arrive at $f_Z(z)$? $\endgroup$ Commented May 13, 2021 at 21:48
  • $\begingroup$ @AaronHendrickson Oh I guess based on leonbloy's solution, I could do the same $\endgroup$ Commented May 13, 2021 at 21:48
  • $\begingroup$ I'm not saying there are the same. What I am saying is that the derivation for both formula are with the exception that in the independent case you can factor the joint density at the end. $\endgroup$ Commented May 13, 2021 at 21:50
  • $\begingroup$ I am surprized that this question has 5 answers whereas it is addressed in many probability textbooks or on MathsSE here for example with 2 nice short proofs by Did. $\endgroup$
    – Jean Marie
    Commented May 13, 2021 at 22:55

4 Answers 4

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From the cumulative distribution function: $$F_Z(z)= P(Z\le z ) = P(X+Y \le z) = P(Y \le z -X) = \int_{-\infty}^{\infty} \int_{-\infty}^{z-x} f_{XY}(x,y)\, dy\, dx $$

Now, $$f_Z(z) = \frac{d F_Z(z) }{ d z} = \int_{-\infty}^{\infty} \frac{d }{ d z} \left( \int_{-\infty}^{z-x} f_{XY}(x,y) dy\, \right) dx = \int_{-\infty}^{\infty} f_{XY}(x,z-x) dx $$

If $X$ and $Y$ are independent we can further write

$$f_Z(z) = \int_{-\infty}^{\infty} f_X(x) f_Y(z-x) dx $$

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    $\begingroup$ My calculus is a little rusty. Could you show the derivative step? Did you just use the fundamental theorem of calculus? $\endgroup$ Commented May 13, 2021 at 21:50
  • $\begingroup$ Yes, the derivative goes inside the outer integral, and then you use en.wikipedia.org/wiki/Leibniz_integral_rule $\endgroup$
    – leonbloy
    Commented May 13, 2021 at 22:56
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The characteristic function of $Z=X+Y$ is $$E[e^{iuZ}]=E[e^{iu(X+Y)}]=\int_{\mathbb{R}^2} e^{iu(x+y)}p_{X,Y}(x,y)dxdy$$ The pdf is found by inverse Fourier Transform. By exchanging integrals $$f_{X+Y}(z)=\frac{1}{2\pi}\int_{\mathbb{R}^2}\bigg(\int_{\mathbb{R}} e^{iu(x+y)}e^{-iuz}du\bigg)p_{X,Y}(x,y)dxdy=$$ $$=\int_\mathbb{R^2}\delta(z-(x+y))p_{X,Y}(x,y)dxdy$$ By symmetry of the Dirac delta $\delta(w)=\delta(-w)$ $$\int_\mathbb{R^2}\delta((x+y)-z)p_{X,Y}(x,y)dxdy=$$ $$(x+y)-z=0 \implies y=z-x $$ $$=\int_\mathbb{R}p_{X,Y}(x,z-x)dx$$


$$P(Z\leq z)=P(X+Y\leq z)=\int P(X+y\leq z|Y=y)p_Y(y)dy=$$ $$=\int_{\infty}^\infty\int^{z-y}_{-\infty}p_{X|Y}(x,y)p_y(y)dxdy=\int_{\infty}^\infty\int^{z-y}_{-\infty}p_{X,Y}(x,y)dxdy$$ $$f_Z(z)=\frac{d}{dz}P(Z\leq z)=\int_{-\infty}^\infty p_{X,Y}(z-y,y)dy$$ By using Leibniz integral rule.

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  • $\begingroup$ This is more involved than I anticipated. Is there a way to just derive this from conditional distribution? $\endgroup$ Commented May 13, 2021 at 21:42
  • $\begingroup$ added @student010101 $\endgroup$
    – Snoop
    Commented May 13, 2021 at 21:51
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There is a property for linear combinations that says if X, Y have joint pdf $f(x, y)$ and $Z=aX+bY+c$, then Z has pdf $$g(z)=\int_{-\infty}^\infty f\left(x, \frac{z-c-ax}b\right)\frac1{|b|}dx$$

For $Z=X+Y$, $$\begin{split}G(z)&=\int_{-\infty}^\infty\int_{-\infty}^{z-x}f(x,y)dydx\end{split}$$

Change of variables $w=x+y$

$$\begin{split}G(z)&=\int_{-\infty}^\infty\int_{-\infty}^{z}f(x,w-x)dwdx\\ &=\int_{-\infty}^z\int_{-\infty}^{\infty}f(x,w-x)dxdw\end{split}$$

Derivative with respect to $z$

$$\begin{split}g(z)&=\left[\frac{d}{dz}(z)\right]\int_{-\infty}^\infty f(x, z-x)dx-0\\ &=1\cdot\int_{-\infty}^\infty f(x, z-x)dx=\int_{-\infty}^\infty f(x, z-x)dx\end{split}$$

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  • $\begingroup$ But how did this come about? $\endgroup$ Commented May 13, 2021 at 21:40
  • $\begingroup$ Would you like it for the general form or just for X+Y? $\endgroup$
    – Vons
    Commented May 13, 2021 at 21:41
  • $\begingroup$ I think just $X + Y$ $\endgroup$ Commented May 13, 2021 at 21:41
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A pdf for a random vector $X:\Omega\to\Bbb R^n$ is any function $g:\Bbb R^n\to \Bbb R$ such that $E[h(X)]=\int_{\Bbb R^n}h(x)g(x)\,dx$ for, say, all bounded summable Borel functions $h:\Bbb R^n\to\Bbb R$.

So, if your random vector is $V=(X,Y)$, then, without too many ceremonies, you can use the change of variables $(t,s)= (x,y+x)$ and Fubini-Tonelli in the integral $$E[h(X+Y)]=\int_{\Bbb R^2} h(x+y)f_{X,Y}(x,y)\,dxdy=\int_{\Bbb R^2}h(s)f_{X,Y}(t,s-t)\,dsdt=\\=\int_{-\infty}^\infty f(s)\left(\int_{-\infty}^\infty f_{X,Y}(t,s-t)\,dt\right)\,ds$$

And there you have it.

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