2.3 Sums and Linear Functions
Result 2.3.1. If \(Y = a + bX\) with \(a\) and \(b\) constants, then \[G_Y(t) = t^{a}\,G_X\!\left (t^{b}\right ).\]
Proof. \[G_Y(t) = E\left (t^{a+bX}\right ) = t^{a}E\left (t^{bX}\right ) = t^{a}E\left [\left (t^{b}\right )^{X}\right ] = t^{a}G_X\!\left (t^{b}\right ),\] the constant \(t^{a}\) coming outside the expectation because it does not depend on \(X\). □
Result 2.3.2. Let \(X_1,\dots ,X_n\) be independent non-negative integer-valued random variables and let \(c_1,\dots ,c_n\) be constants. Then \[G_{\sum _{i=1}^{n}c_iX_i}(t) = \prod _{i=1}^{n} G_{X_i}\!\left (t^{c_i}\right ).\]
Proof. Put \(Y = \sum _{i=1}^{n}c_iX_i\). Then \[G_Y(t) = E\left (t^{Y}\right ) = E\left (t^{c_1X_1 + c_2X_2 + \cdots + c_nX_n}\right ) = E\left (t^{c_1X_1}\,t^{c_2X_2}\cdots t^{c_nX_n}\right ).\] The variables are independent, so the expectation of the product is the product of the expectations: \[G_Y(t) = \prod _{i=1}^{n}E\left (t^{c_iX_i}\right ) = \prod _{i=1}^{n}E\left [\left (t^{c_i}\right )^{X_i}\right ] = \prod _{i=1}^{n}G_{X_i}\!\left (t^{c_i}\right ).\] □
Remark. Taking every \(c_i = 1\) gives the case that matters most: \[G_{X_1+\cdots +X_n}(t) = \prod _{i=1}^{n}G_{X_i}(t).\] Convolution has become multiplication. This is the reason generating functions are introduced at all, and the next section makes the point concrete: it is far easier to multiply two polynomials than to evaluate \(P(X+Y=k) = \sum _j P(X=j)P(Y=k-j)\) term by term.
Independence is essential and is the only place it is used. For dependent variables the expectation of the product is not the product of the expectations, and the result fails.
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