Mean and Variance of exponential function

Question

Given this function with j and k as unknown parameters. What is the expression of Variance and Mean of this exponential function?

$$f_{j,k}(y)=\frac{\sqrt{j}}{\sqrt{2 \pi}}e^{\sqrt{jk}}y^{-\frac{1}{2}} \text{exp}\left( -\frac{1}{2} (j y + \frac{k}{y}) \right) \quad \quad y>0$$

Answer 1

I put your question into Mathematica and got the following expression for the mean:

$$ \frac{1}{2 j \sqrt{2 \pi }}e^{\sqrt{jk}} \left(-4 e^{-\frac{k+j x^2}{2 x}} \sqrt{j} \sqrt{x}+e^{-\sqrt{j} \sqrt{k}} \sqrt{2 \pi } \left(\left(1+\sqrt{j} \sqrt{k}\right) \text{Erf}\left[\frac{-\sqrt{k}+\sqrt{j} x}{\sqrt{2} \sqrt{x}}\right]-\left(-1+\sqrt{j} \sqrt{k}\right) \left(1-e^{2 \sqrt{j} \sqrt{k}}+e^{2 \sqrt{j} \sqrt{k}} \text{Erf}\left[\frac{\sqrt{k}+\sqrt{j} x}{\sqrt{2} \sqrt{x}}\right]\right)\right)\right) $$

I'm not sure how easy it would be to derive that...

[edited for math error]

Answer 2

As I show you in this post Show that a statistical model belongs to exponential family the parametrization is $$ f_{\pmb{\theta}}(y) =h(y)g(\pmb{\theta}) \exp\left(\pmb{\eta}(\pmb{\theta})\cdot \pmb{T}(x)) \right)= h(y) \exp\left(\pmb{\eta}(\pmb{\theta})\cdot \pmb{T}(x) - A(\pmb{\theta}) \right) $$ with $$ \begin{align} \pmb{\theta}&=(j,k)\\ h(y)&=\frac{1}{\sqrt{2 \pi}}y^{-\frac{1}{2}}\\ g(\pmb{\theta})&=g(j,k)=\sqrt{j}\,\operatorname{e}^{\sqrt{jk}}=\operatorname{e}^{-A(\pmb\theta)}\\ \pmb{\eta}(\pmb{\theta})&=\left(\eta_1(j,k),\eta_2(j,k)\right)=(j,k)=\pmb{\pmb\theta}\\ \pmb T(y)&=\left(T_1(y),T_2(y)\right)=\left(-\frac{y}{2},-\frac{1}{2y}\right) \end{align} $$ and $A(\pmb{\theta})=-\log(g(\pmb{\theta}))=A(\pmb{\eta})=-\frac{1}{2}\log j-\sqrt{jk}$.

The cumulant generating function of the sufficient statistic $\pmb T$ is $$ K(\pmb t) = A(\pmb\eta+\pmb t) - A(\pmb\eta) $$ so that $$\begin{align} \Bbb E(T_{\nu})&=\left.\frac{\partial K(\pmb t)}{\partial t_{\nu}}\right|_{\pmb t=0}=\frac{\partial A(\pmb \eta)}{\partial \eta_{\nu}}\\ \operatorname{Var}(T_{\nu})&=\left.\frac{\partial^2 K(\pmb t)}{\partial t_{\nu}^2}\right|_{\pmb t=0}=\frac{\partial^2 A(\pmb \eta)}{\partial \eta_{\nu}^2} \end{align} $$ So we have $$\begin{align} \Bbb E(T_{1})&=\frac{\partial A(j,k)}{\partial j}=-\frac{1}{2}\left(\frac{1}{j}+\sqrt{\frac{k}{j}}\right)\\ \operatorname{Var}(T_{1})&=\frac{\partial^2 A(j,k)}{\partial j^2}=\frac{1}{2}\left(\frac{1}{j^2}+\frac{1}{2}\sqrt{\frac{k}{j^3}}\right) \end{align} $$ and finally $$\begin{align} \Bbb E(Y)&=-2\Bbb E(T_{1})=\frac{1}{j}+\sqrt{\frac{k}{j}}\\ \operatorname{Var}(Y)&=4\operatorname{Var}(T_{1})=\frac{2}{j^2}+\sqrt{\frac{k}{j^3}} \end{align} $$