TheoremComplete

Joint and Conditional Distributions - Main Theorem

The convolution formula provides a method for computing the distribution of sums of independent random variables.

Convolution Theorem

Theorem

If XX and YY are independent continuous random variables with PDFs fXf_X and fYf_Y, then Z=X+YZ = X + Y has PDF: fZ(z)=fX(x)fY(zx)dx=fY(y)fX(zy)dyf_Z(z) = \int_{-\infty}^{\infty} f_X(x) f_Y(z-x) \, dx = \int_{-\infty}^{\infty} f_Y(y) f_X(z-y) \, dy

This is the convolution of fXf_X and fYf_Y, written fZ=fXfYf_Z = f_X * f_Y.

Proof: For any zz: FZ(z)=P(X+Yz)=x+yzfX,Y(x,y)dxdyF_Z(z) = P(X + Y \leq z) = \int\int_{x+y \leq z} f_{X,Y}(x,y) \, dx \, dy

By independence, fX,Y(x,y)=fX(x)fY(y)f_{X,Y}(x,y) = f_X(x)f_Y(y): =fX(x)[zxfY(y)dy]dx=fX(x)FY(zx)dx= \int_{-\infty}^{\infty} f_X(x) \left[\int_{-\infty}^{z-x} f_Y(y) \, dy\right] dx = \int_{-\infty}^{\infty} f_X(x) F_Y(z-x) \, dx

Differentiating with respect to zz: fZ(z)=fX(x)fY(zx)dxf_Z(z) = \int_{-\infty}^{\infty} f_X(x) f_Y(z-x) \, dx

Example

Sum of Two Uniforms: X,YUniform(0,1)X, Y \sim \text{Uniform}(0,1) independent. Find PDF of Z=X+YZ = X + Y.

fZ(z)=fX(x)fY(zx)dxf_Z(z) = \int_{-\infty}^{\infty} f_X(x) f_Y(z-x) \, dx

For 0<z<10 < z < 1: fX(x)=1f_X(x) = 1 for 0<x<z0 < x < z, fY(zx)=1f_Y(z-x) = 1 for 0<zx<10 < z-x < 1: fZ(z)=0z11dx=zf_Z(z) = \int_0^z 1 \cdot 1 \, dx = z

For 1<z<21 < z < 2: Both conditions give z1<x<1z-1 < x < 1: fZ(z)=z1111dx=2zf_Z(z) = \int_{z-1}^1 1 \cdot 1 \, dx = 2 - z

Result: fZ(z)={z0<z<12z1<z<2f_Z(z) = \begin{cases} z & 0 < z < 1 \\ 2-z & 1 < z < 2 \end{cases} (triangular distribution)

Discrete Convolution

For discrete independent XX and YY: pZ(z)=xpX(x)pY(zx)=ypY(y)pX(zy)p_Z(z) = \sum_x p_X(x) p_Y(z-x) = \sum_y p_Y(y) p_X(z-y)

Example

Sum of two dice: X,Y{1,2,3,4,5,6}X, Y \in \{1,2,3,4,5,6\} with pX(k)=pY(k)=1/6p_X(k) = p_Y(k) = 1/6.

For Z=X+Y=7Z = X + Y = 7: pZ(7)=k=16pX(k)pY(7k)=k=1616161{17k6}p_Z(7) = \sum_{k=1}^6 p_X(k) p_Y(7-k) = \sum_{k=1}^6 \frac{1}{6} \cdot \frac{1}{6} \cdot \mathbb{1}\{1 \leq 7-k \leq 6\} =6×136=16= 6 \times \frac{1}{36} = \frac{1}{6}

Six ways: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).

Remark

Convolution is computationally intensive but conceptually fundamental. For practical calculations with many sums, MGFs or characteristic functions are often more efficient. However, convolution provides geometric intuition: we're "sliding" one PDF along the other and integrating overlaps.