I need to compute the moment-generating function of the non-central chi-squared distribution, but I have no idea where to begin.


1 Answer 1


Let $Z$ have a standard normal distribution, with mean $0$ and variance $1$, then $(Z+\mu)^2$ has a noncentral chi-squared distribution with one degree of freedom. The moment-generating function of $(Z+\mu)^2$ then is \begin{equation} E[e^{t(Z+\mu)^2}] = \int_{-\infty}^{+\infty} e^{t(z+\mu)^2} f_Z(z) \text{d}z \end{equation} with $f_Z(z) = \frac{1}{\sqrt{2\pi}}e^{-z^2/2}$ the density of $Z$. Then, \begin{align*} E[e^{t(Z+\mu)^2}] & = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{t(z+\mu)^2} e^{-z^2/2}\text{d}z\\ & = \frac{1}{\sqrt{2\pi}} \int \exp[-(\frac{1}{2}-t)z^2+2\mu t z +\mu^2 t] \text{d} z\\ & = \frac{1}{\sqrt{2\pi}} \int \exp[-(\frac{1}{2}-t) (z-Q)^2+\mu^2 t + \frac{2\mu^2 t^2}{1-2t}] \text{d} z \qquad\text{with } Q=\frac{2\mu t}{1-2t}\\ & = \exp[\mu^2 t + \frac{2\mu^2t^2}{1-2t}] \frac{1}{\sqrt{2\pi}} \int \exp[-\frac{(z-Q)^2}{2(1-2t)^{-1}}] \text{d}z\\ & = \exp[\mu^2 t + \frac{2\mu^2t^2}{1-2t}] (1-2t)^{-1/2} \times \frac{1}{\sqrt{2\pi}(1-2t)^{-1/2}} \int \exp[-\frac{(z-Q)^2}{2(1-2t)^{-1}}] \text{d}z\\ & = \exp[\mu^2 t + \frac{2\mu^2t^2}{1-2t}] (1-2t)^{-1/2} \times 1\\ & = \exp[\frac{\mu^2 t}{1-2t} ] (1-2t)^{-1/2} \end{align*} By definition, a noncentral chi-squared distributed random variable $\chi^2_{n,\lambda}$ with $n$ df and parameter $\lambda=\sum_{i=1}^n \mu_i^2$ is the sum of the squares of $n$ independent normal variables $X_i=Z_i+\mu_i$, $i=1,\ldots,n$. That is, \begin{equation*} \chi^2_{n,\lambda} = \sum_{i=1}^n X_i^2 = \sum_{i=1}^n (Z_i+\mu_i)^2 \end{equation*} Because all terms are jointly independent and using the above result, we have \begin{align*} E[e^{t\chi^2_{n,\lambda}}] & = E[\prod_i \exp[t(Z_i+\mu_i)^2]]\\ & = \prod_i E[\exp[t(Z_i+\mu_i)^2]]\\ & = \prod_i (1-2t)^{-1/2} \exp[\frac{\mu_i^2t}{1-2t}]\\ & = (1-2t)^{-n/2} \exp[\frac{t}{1-2t}\sum_i \mu_i^2]\\ & = (1-2t)^{-n/2} \exp[\frac{\lambda t}{1-2t}]\\ \end{align*}

  • $\begingroup$ Hi Stijn, I don't understand how you have $E[e^{t(Z+\mu)^2}] = \int_{-\infty}^{+\infty} e^{t(z+\mu)^2} f_Z(z)$, with $f_Z(z) = \dfrac{1}{\sqrt{2\pi}}e^-{z^2/2}.$ $X = (Z+\mu)^2$ has a noncentral $\chi^2$ distribution, so shouldn't we have a moment generating function of the form $E[e^{t(Z + \mu)^2}] = \int e^{t(Z + \mu)^2} f_X(x)dx$, where $f_X(x)$ is the probability density function of the noncentral $\chi^2$ distribution? $\endgroup$
    – titusAdam
    Mar 6, 2018 at 17:21
  • $\begingroup$ @titusAdam: No. The correct formula for the expectation of a function $g(Z)$ of a (continuous) random variable $Z$ with density function $f_Z(z)$ is $\text{E}[g(Z)]=\int g(z) f_Z(z)\text{d}z$. In our case here we have $\text{E}[e^{tX}]=\int e^{tx}f_X(x)\text{d}x=\text{E}[e^{t(Z+\mu)^2}]=\int e^{t(z+\mu)^2}f_Z(z)\text{d}z$. So either you integrate over $x$ or over $z$, but not something in between. $\endgroup$ Mar 8, 2018 at 19:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.