Let $X_1,….X_n$ be a random sample from the inverse Gaussian distribution with pdf
$$f(x|\mu,\lambda)=\left(\frac{\lambda}{2\pi x^3}\right)^{1/2} \exp\left(\frac{-\lambda (x-\mu)^2}{2 \mu^2 x}\right),\quad0 \lt x \lt \infty$$
For $n=2$, show that $\overline X$ has an inverse Gaussian distribution, $n\lambda/T$ has a $\chi^2_{n-1}$ distribution, and they are independent.
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Best Answer
Let $\mathcal{K}_X(t)=\log\left(\Bbb{E}\left(\operatorname{e}^{tX}\right)\right)$ the cumulant generating function of $X\sim \mathcal{IG}(\mu,\lambda)$.
The pdf of $X\sim \mathcal{IG}(\mu,\lambda)$ is $$ \begin{align} f(x;\mu,\lambda) &= \left(\frac{\lambda}{2 \pi x^3}\right)^{1/2} \exp\left({\frac{-\lambda (x-\mu)^2}{2 \mu^2 x}} \right)\qquad x>0\tag 1 \end{align} $$ that can be rewritten as $$ \begin{align} f(x;\mu,\lambda) &= \left(\frac{\lambda}{2 \pi x^3}\right)^{1/2} \exp\left({-\frac{\lambda}{2\mu^2} x+\frac{\lambda}{\mu}-\frac{\lambda}{2x}} \right)\qquad x>0\tag 2 \end{align} $$ so the cumulant generating function, using the (2), is $$ \begin{align} \mathcal{K}_X(t)&=\log\left(\Bbb{E}\left(\operatorname{e}^{tX}\right)\right)=\log\left(\int_0^{+\infty}\operatorname{e}^{tX}f(x;\mu,\lambda)\operatorname{d}x\right)\\ &= \frac{\lambda}{\mu}+\log\left(\int_0^{+\infty}\exp\left({-\frac{\lambda}{2\mu^2} x+tx-\frac{\lambda}{2x}} \right)\left(\frac{\lambda}{2 \pi x^3}\right)^{1/2} \operatorname{d}x\right)\\ &=\frac{\lambda}{\mu}+\log\left(\underbrace{\int_0^{+\infty}\exp\left({-\left[\frac{1}{2\mu^2}-\frac{t}{\lambda}\right]\lambda x-\frac{\lambda}{2x}} \right)\left(\frac{\lambda}{2 \pi x^3}\right)^{1/2} \operatorname{d}x}_{\mathcal{I}}\right) \end{align} $$ Putting $\beta^2=2\lambda^2\left(\frac{1}{2\mu^2}-\frac{t}{\lambda}\right)$ and substituting $y=\left(\frac{1}{2\mu^2}-\frac{t}{\lambda}\right)\lambda x=\frac{\beta^2}{2\lambda}x$, the integral $\mathcal{I}$ becomes $$ \mathcal{I}=\frac{\beta}{2\sqrt \pi}\int_0^{+\infty}\exp\left({-y-\frac{\beta^2}{4y}}\right) \frac{\operatorname{d}y}{y^{\frac{1}{2}+1}} $$ and recalling the integral representation of the modified Bessel function $K_{\nu}(z)$ $$ K_{\nu}(z)=\frac{1}{2}\left(\frac{z}{2}\right)^{\nu}\int_0^{+\infty}\exp\left({-t-\frac{z^2}{4t}}\right) \frac{\operatorname{d}t}{t^{\nu+1}} $$ we have $$ \mathcal{I}=\frac{\beta}{2\sqrt \pi}\cdot 2\left(\frac{2}{\beta}\right)^{\frac{1}{2}}K_{\frac{1}{2}}(\beta)=\operatorname{e}^{-\beta} $$
using the identity $K_\frac{1}{2}(z)=\left(\frac{\pi}{2z}\right)^{\frac{1}{2}} \mathrm{e}^{-z},\, z>0$.
Thus we have $$ \begin{align} \mathcal{K}_X(t)&=\frac{\lambda}{\mu}+\log (\operatorname{e}^{-\beta})=\frac{\lambda}{\mu}-\left[2\lambda^2\left(\frac{1}{2\mu^2}-\frac{t}{\lambda}\right)\right]^{\frac{1}{2}} \end{align} $$ and finally $$ \mathcal{K}_X(t)=\frac{\lambda}{\mu}\left[1-\left(1-\frac{2\mu^2 t}{\lambda}\right)^{\frac{1}{2}}\right].\tag 3 $$ Observing that if $X\sim \mathcal{IG}(\mu,\lambda)$, then $\frac{X}{n}\sim \mathcal{IG}\left(\frac{\mu}{n},\frac{\lambda}{n}\right)$ and if $X_1,\ldots,X_n$ aree iid $\mathcal{IG}(\mu,\lambda)$ then the cumulant generating function of $n\bar{X}=\sum_{i=1}^n X_i$ is the sum of the cumulant generating of each $X_i$: $$ \mathcal{K}_{\bar{X}}(t)=\sum_{i=1}^n\mathcal{K}_{X_i/n}(t)=\frac{n\lambda}{\mu}\left[1-\left(1-\frac{2\mu^2 t}{n\lambda}\right)^{\frac{1}{2}}\right]\tag 4 $$ that is the cumulant generating function of an inverse gaussian distribution with mean $\mu$ and shape $n\lambda$: $$ \bar{X}=\frac{1}{n}\sum_{i=1}^n X_i \sim \mathcal{IG}(\mu,n\lambda).\tag 5 $$
Recall a useful result for exponential families.
If $X_1,\ldots, X_n$ are independently and identically distributed (iid) with the common density $$ p(x;\theta, \eta) = h(x, \eta) \operatorname{e}^{\theta T(x) − A(\theta, \eta)}\tag 6 $$ and if the density of $U=U(X_1,\ldots, X_n)$ is given by $$ q(u;\theta, \eta) = h^*(u, \eta) \operatorname{e}^{n\theta T(u) − A^*(\theta, \eta)}\tag 7 $$ for some $A^∗$ and $h^∗$, then $U$ is independent of $V = \sum_{i=1}^n T(X_i)−nT(U)$.
Observing that if $X\sim \mathcal{IG}(\mu,\lambda)$, then $Y=\frac{X}{\mu}\sim \mathcal{IG}\left(\frac{\mu}{\mu},\frac{\lambda}{\mu}\right)=\mathcal{IG}\left(1,\phi\right)$ with $\phi=\frac{\lambda}{\mu}$ and the pdf of $\frac{X}{\mu}$ is $$ \begin{align} f(y;1,\phi) &= \left(\frac{\phi}{2 \pi y^3}\right)^{1/2} \exp\left({-\frac{\phi}{2}\left(y+\frac{1}{y}\right)+\phi} \right)\qquad u>0\tag 2 \end{align} $$ which is of form (6) with $\theta=-\frac{\phi}{2}$ and $T(y)=y+\frac{1}{y}$. Using (5), and putting $Y_i=\frac{X_i}{\mu}$ for $i=1,\ldots,n$, we know that $$ \bar{Y}=\frac{1}{n}\sum_{i=1}^n Y_i \sim \mathcal{IG}(1,n\phi) $$ with a density of form (7).
Hence, $\bar{Y}=U(X_1,\ldots,X_n)$ is independent of $$V= \sum_{i=1}^n \left(Y_i+\frac{1}{Y_i}\right)-n\left(\bar{Y}+\frac{1}{\bar{Y}}\right)=\sum_{i=1}^n \frac{1}{Y_i}-\frac{n}{\bar{Y}} $$ and finally $\bar{X}=\mu\bar{Y}$ is independent of $Z=\frac{V}{\mu}$, that is $$ \bar{X}=\frac{1}{n}\sum_{i=1}^n X_i \qquad\text{is independent of }\qquad Z=\sum_{i=1}^n \left(\frac{1}{X_i}-\frac{1}{\bar{X}}\right) $$ Note that $\hat{\mu}=\bar{X}$ and $\hat{\lambda}=\frac{n}{Z}$ are the maximum likelihood estimators (MLE) of $\mu$ and $\lambda$.
Of course, there are other ways of proving these independence results; among which, Basu’s theorem is the most popular. Simply put, this theorem says: if $T$ is a complete sufficient statistic for a family of probability distributions $\mathcal{P}$, and $S$ is an ancillary statistic, then $T$ and $S$ are independent.
As $X_1,\,\ldots,X_n$ are independent, the cumulative density $f(X_1,\,\ldots,X_n)$ is the product of the marginal densities of $X_1,\,\ldots,X_n$. Observe that $$ \Bbb{E}(g(X_1,\dots,X_n)) = \int_{-\infty}^\infty\cdots \int_{-\infty}^\infty g(x_1,\ldots,x_n)~f(x_1,\ldots,x_n)~\mathrm{d}x_1\cdots \mathrm{d}x_n . $$
The moment generating function of $Y=\frac{Z}{n}=$ is then $$ \begin{align} \Bbb{E}\left(\operatorname{e}^{tY}\right)&=\idotsint_{\text{all }x_i>0}\operatorname{e}^{tY}\left(\prod_{i=1}^n f(x_i;\mu,\lambda)\operatorname{d}x_i\right) \end{align} $$ Observing that $$ \begin{align} \prod_{i=1}^n f(x_i;\mu,\lambda)&=\left[\prod_{i=1}^n \left(\frac{\lambda}{2 \pi x_i^3}\right)^{1/2}\right] \exp\left({-\frac{\lambda}{2\mu^2}\sum_{i=1}^n x_i+\sum_{i=1}^n\frac{\lambda}{\mu}-\sum_{i=1}^n\frac{\lambda}{2x_i}} \right)\\ &=\left(\frac{\lambda}{2 \pi}\right)^{n/2}\left[\prod_{i=1}^n x_i^{-3/2} \right] \exp\left(-\frac{n\lambda}{2\mu^2}\bar{x}+\frac{n\lambda}{\mu}-\frac{n\lambda}{2\bar{x}}- \frac{n\lambda}{2}y\right)\\ &=\left(\frac{\lambda}{2 \pi}\right)^{\frac{n-1}{2}}\bar{x}^{-3/2}n^{-1/2}f(\bar{x};\mu,n\lambda)\operatorname{e}^{-\frac{n\lambda}{2}y} \end{align} $$ we have $$ \begin{align} \Bbb{E}\left(\operatorname{e}^{tY}\right)&=\idotsint_{\text{all }x_i>0}\operatorname{e}^{\left(t-\frac{n\lambda}{2}\right)y}\left(\frac{\lambda}{2 \pi}\right)^{\frac{n-1}{2}}\bar{x}^{-3/2}n^{-1/2}f(\bar{x};\mu,n\lambda)\left(\prod_{i=1}^n x_i^{-3/2}\operatorname{d}x_i\right)\\ &=\int_{0}^{\infty}f(\bar{x};\mu,n\lambda)\idotsint_{\bar{x}\text{ is constant}}\operatorname{e}^{\left(t-\frac{n\lambda}{2}\right)y}\left(\frac{\lambda}{2 \pi}\right)^{\frac{n-1}{2}}\bar{x}^{-3/2}n^{-1/2}\left(\prod_{i=1}^n x_i^{-3/2}\operatorname{d}x_i\right). \end{align} $$ Form the relation $\Bbb{E}\left(U\right)=\Bbb{E}_{V}\left(\Bbb{E}\left(\left.U\right|V\right)\right)$, it follows taht the conditional moment generating function of $Y$ for fixed $\bar{X}$ can be found taking the partial derivative with respect to $\bar x$ of the multiple integral: $$ \begin{align} \Bbb{E}\left(\left.\operatorname{e}^{tY}\right|\bar{X}\right) &=\frac{\partial}{\partial\bar x}\underbrace{\idotsint_{\bar{x}\text{ is constant}}\operatorname{e}^{\left(t-\frac{n\lambda}{2}\right)y}\left(\frac{\lambda}{2 \pi}\right)^{\frac{n-1}{2}}\bar{x}^{-3/2}n^{-1/2}\left(\prod_{i=1}^n x_i^{-3/2}\operatorname{d}x_i\right)}_{\mathcal{J}(t)}. \end{align} $$ To evaluate the integral, we take first $t=0$, obtaining $$ \frac{\partial}{\partial\bar x}\mathcal{J}(0)=n^{-1/2}\left(\frac{2 \pi}{\lambda}\right)^{\frac{n-1}{2}}. $$ By substituting $\lambda-\frac{2t}{n}$ we find $$ \Bbb{E}\left(\left.\operatorname{e}^{tY}\right|\bar{X}\right)=\left(1-\frac{2t}{n\lambda}\right)^{-\frac{n-1}{2}} $$ wich is the moment generating function of a chi square variable with $n-1$ degree of freedom: $$Y\sim \frac{1}{n\lambda}\chi^2_{n-1}.$$
Finally we have $$ \frac{n\lambda}{\sum_{i=1}^n \left(\frac{1}{X_i}-\frac{1}{\bar{X}}\right)}\sim \chi^2_{n-1}. $$