Chi-squared Distribution: Expectation, Variance and Moment Generating Function Proofs

Expectation

\[\begin{align*} E(X) &= \int_0^\infty x \frac{1}{2^{k/2}\Gamma(\frac{k}{2})} x^{\frac{k}{2}-1} e^{\frac{-x}{2}} dx \\ &= \frac{1}{2^{k/2}\Gamma(\frac{k}{2})} \int_0^\infty x \cdot x^{\frac{k}{2}-1} e^{\frac{-x}{2}} dx \\ &= \frac{1}{2^{k/2}\Gamma(\frac{k}{2})} \int_0^\infty x^{\frac{k}{2}} e^{\frac{-x}{2}} dx \\ & \text{let: } u = \frac{x}{2} \\ \Rightarrow x &= 2u \\ \therefore dx &= 2du \\ \text{then: } x^{\frac{k}{2}} &= (2u)^{\frac{k}{2}} = (2)^{\frac{k}{2}}(u)^{\frac{k}{2}} \\ E(X) &= \frac{1}{2^{k/2}\Gamma(\frac{k}{2})} \int_0^\infty (2)^{\frac{k}{2}}(u)^{\frac{k}{2}} e^{-u} 2du \\ &= \frac{2}{\Gamma(\frac{k}{2})} \int_0^\infty u^{\frac{k}{2}} e^{-u} du \\ &= \frac{2}{\Gamma(\frac{k}{2})} \int_0^\infty u^{(\frac{k}{2}+1)-1} e^{-u} du \\ &= \frac{2}{\Gamma(\frac{k}{2})} \Gamma(\frac{k}{2}+1) \\ &= \frac{2}{\Gamma(\frac{k}{2})} \frac{k}{2} \Gamma(\frac{k}{2}) \\ &= k \end{align*}\]

Variance

\[\begin{align*} E(X^2) &= \int_0^\infty x^2 \frac{1}{2^{k/2}\Gamma(\frac{k}{2})} x^{\frac{k}{2}-1} e^{\frac{-x}{2}} dx \\ &= \frac{1}{2^{k/2}\Gamma(\frac{k}{2})} \int_0^\infty x^2 \cdot x^{\frac{k}{2}-1} e^{\frac{-x}{2}} dx \\ &= \frac{1}{2^{k/2}\Gamma(\frac{k}{2})} \int_0^\infty x^{\frac{k}{2}+1} e^{\frac{-x}{2}} dx \\ & \text{let: } u = \frac{x}{2} \\ \Rightarrow x &= 2u \\ \therefore dx &= 2du \\ \text{then: } x^{\frac{k}{2}+1} &= (2u)^{\frac{k}{2}+1} = (2)^{\frac{k}{2}+1}(u)^{\frac{k}{2}+1} \\ E(X^2) &= \frac{1}{2^{k/2}\Gamma(\frac{k}{2})} \int_0^\infty (2)^{\frac{k}{2}+1}(u)^{\frac{k}{2}+1} e^{-u} 2du \\ &= \frac{4}{\Gamma(\frac{k}{2})} \int_0^\infty u^{\frac{k}{2}+1} e^{-u} du \\ &= \frac{4}{\Gamma(\frac{k}{2})} \int_0^\infty u^{(\frac{k}{2}+2)-1} e^{-u} du \\ &= \frac{4}{\Gamma(\frac{k}{2})} \Gamma(\frac{k}{2}+2) \\ &= \frac{4}{\Gamma(\frac{k}{2})} \frac{k}{2} (\frac{k}{2}+1) \Gamma(\frac{k}{2}) \\ &= 4 \cdot \left(\frac{k^2}{4} + \frac{k}{2}\right) \\ &= k^2 + 2k \\ Var(X) &= k^2 + 2k - (k)^2 \\ &= 2k \end{align*}\]

Moment Generating Function

\[\begin{align*} M_X(t) &= \int_0^\infty e^{tx} \frac{1}{2^{\frac{k}{2}}\Gamma(\frac{k}{2})} x^{\frac{k}{2}-1}e^{-\frac{x}{2}}dx \\ &= \frac{1}{2^{\frac{k}{2}}\Gamma(\frac{k}{2})} \int_0^\infty x^{\frac{k}{2}-1}e^{tx -\frac{x}{2}}dx \\ &= \frac{1}{2^{\frac{k}{2}}\Gamma(\frac{k}{2})} \int_0^\infty x^{\frac{k}{2}-1}e^{-(\frac{1}{2} - t)x}dx \\ \end{align*}\]

Note:

• if \(\frac{1}{2} - t <0\), then \(e^{-(\frac{1}{2}-t)x} \to \infty\)
• if \(\frac{1}{2} - t =0\), then \(e^{-(\frac{1}{2}-t)x} = 1\); this results in \(\int_0^\infty x^{(n/2)-1} dx\). However, as \(n\) is strictly positive, this integral diverges
• Therefore, we must have \(\frac{1}{2} - t > 0 \Rightarrow t < \frac{1}{2}\)

\[\begin{align*} & \text{Let: } u = \left(\frac{1}{2} - t \right)x \Rightarrow x = \frac{u}{\frac{1}{2} - t} \text{ and } dx = \frac{1}{\frac{1}{2} - t} \\ &= \frac{1}{2^{\frac{k}{2}}\Gamma(\frac{k}{2})(\frac{1}{2} - t)} \int_0^\infty \left(\frac{u}{\frac{1}{2} - t}\right)^{\frac{n}{2}-1} e^{-u}du \\ &= \frac{1}{2^{\frac{k}{2}}\Gamma(\frac{k}{2})(\frac{1}{2} - t)^{\frac{n}{2}}} \int_0^\infty u^{\frac{n}{2}-1} e^{-u}du \\ &= \frac{\Gamma(\frac{n}{2})}{\Gamma(\frac{n}{2})} (1-2t)^{-\frac{n}{2}} \\ &= (1-2t)^{-\frac{n}{2}} \end{align*}\]