1.2 Sums of independent random variables
Let \(X\) and \(Y\) be independent continuous random variables, and let \(Z = X + Y\). Then \begin {align*} F_Z(z) &= P(Z \le z) = P(X + Y \le z) \\ &= \iint \limits _{x+y\le z}f_{X,Y}(x, y)\,dx\,dy \\ &= \iint \limits _{x+y\le z}f_X(x)f_Y(y)\,dx\,dy \\ &= \int _{-\infty }^{\infty }\int _{-\infty }^{z-y}f_X(x)f_Y(y)\,dx\,dy . \end {align*}
Hence \[ F_Z(z) = \int _{-\infty }^{\infty }F_X(z - y)f_Y(y)\,dy , \] and differentiating with respect to \(z\) gives the convolution formula \[ f_Z(z) = \int _{-\infty }^{\infty }f_X(z - y)f_Y(y)\,dy . \]
Example 1.2.1. Let \(X\) and \(Y\) be independent and uniformly distributed on \((0, 2)\). Find the pdf of \(Z = X + Y\).
Solution. Here \(f_X(x) = f_Y(y) = \frac 12\) for \(x, y \in (0, 2)\). The integrand \(f_X(z - y)f_Y(y)\) is \(\frac 14\) when both \(0 < y < 2\) and \(0 < z - y < 2\), and \(0\) otherwise. If \(0 < z < 2\), this means \(0 < y < z\), so \[ f_Z(z) = \int _0^z\frac 12\cdot \frac 12\,dy = \frac z4 . \] If \(2 < z < 4\), it means \(z - 2 < y < 2\), so \[ f_Z(z) = \int _{z-2}^2\frac 14\,dy = \frac y4\Big |_{z-2}^2 = 1 - \frac z4 . \] Therefore \[ f_Z(z) = \begin {cases} \dfrac z4 , & 0 < z < 2 , \\[1ex] 1 - \dfrac z4 , & 2 < z < 4 , \end {cases} \] a triangular density on \((0, 4)\).
The sum of two independent Poisson variables
Let \(X\) and \(Y\) be independent Poisson random variables with means \(\mu _1\) and \(\mu _2\), and let \(Z = X + Y\). Then \begin {align*} P(Z = z) &= P(X + Y = z) = \sum _{x=0}^zP(X = x, Y = z - x) \\ &= \sum _{x=0}^zP(X = x)P(Y = z - x) \\ &= \sum _{x=0}^z\frac {\mu _1^xe^{-\mu _1}}{x!}\cdot \frac {\mu _2^{z-x}e^{-\mu _2}}{(z - x)!} \\ &= e^{-(\mu _1+\mu _2)}\sum _{x=0}^z\frac {\mu _1^x\mu _2^{z-x}}{x!\,(z - x)!} \\ &= \frac {e^{-(\mu _1+\mu _2)}}{z!}\sum _{x=0}^z\binom zx\mu _1^x\mu _2^{z-x} \\ &= \frac {(\mu _1 + \mu _2)^ze^{-(\mu _1+\mu _2)}}{z!} , \end {align*}
by the binomial theorem. Therefore \(Z \sim P(\mu _1 + \mu _2)\): the sum of independent Poisson variables is Poisson, with the means added.
Note.
- 1.
- If \(X\) and \(Y\) are independent, then \[ M_Z(t) = E\bigl (e^{tZ}\bigr ) = E\bigl (e^{tX + tY}\bigr ) = E\bigl (e^{tX}\bigr )E\bigl (e^{tY}\bigr ) = M_X(t)M_Y(t) . \]
- 2.
- If \(X_1, X_2, \dots , X_n\) are independent Poisson variables with \(E(X_j) = \mu _j\), then \[ Z = \sum _{j=1}^nX_j \sim P\Bigl (\sum _{j=1}^n\mu _j\Bigr ) . \] This follows from the case of two variables by mathematical induction.
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