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I'm trying to understand the intuition behind convolutions for densities Suppose X,Y are RVs and Z = X + Y

then $f_Z(z) = \int_\infty^\infty f(z-y,y)dy$

The lower bound should be negative infinity but I'm having formatting trouble.

Why is this true. By substitution, I feel like this is true: $f_X(x) = \int_\infty^\infty f(z-y,y)dy$

rather than the former

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  • $\begingroup$ is it clear or you need more details or answer like why etc? $\endgroup$ – Seyhmus Güngören Nov 1 '14 at 1:34
  • $\begingroup$ Hey, I appreciate the followup. Of the three listed the only one I understand is the formal proof by mookid. Also, for your proof, how come you can write the probability CDF into your integral in the first line. In the second line, what is regularity conditions? $\endgroup$ – Zhulu Nov 2 '14 at 4:43
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The intuition is the following: let us note $P(X\in dx)$ for the probability that the random variable $X$ is in $[x,x+dx]$.

Then: $$ P(Z\in dz) = \iint_{z < x+y < z+dz} f(x,y) dxdy $$

Now with the change of variable $x =z'-y$: $$ P(Z\in dz) = \iint_{z < z' < z+dz} f(z'-y,y) dz'dy \simeq \left[\int f(z'-y,y) dy\right]\times dz $$

Hence the density of the sum is $$ f_Z(z) = \int f(z'-y,y) dy $$


note that the formal proof is not that different: take $g\ge 0$ measurable.

$$ Eg(Z) = \iint f(x,y)g(x+y) dxdy = \iint f(z-y,y) g(z) dzdy \\=\int \left[ \int f(z-y,y) dy\right] g(z) dz \\\implies f_Z(z) = \int f(z-y,y) dy $$

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$$F_Z(z)=P(Z<z)=P(X+Y<z)=\int P(X<z-y)f_Y(y)\mathrm{d}y=\int F_X(z-y) f_Y(y)\mathrm{d}y$$ $$\frac{\partial F_Z(z)}{\partial\mathrm{d}z}=\frac{\partial \left(\int F_X(z-y) f_Y(y)\mathrm{d}y\right)}{\partial\mathrm{d}z}\overset{\mbox{regularity conditions}}{=} \int \frac{\partial F_X(z-y)}{\partial\mathrm{d}z }f_Y(y)\mathrm{d}y=\int f_X(z-y) f_Y(y)\mathrm{d}y$$

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