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I am supposing that I have a random variable $T$ with pdf

$$ f(t\mid\theta ) = \sqrt{\frac{2}{\pi}}\theta \frac{e^{2t}}{(e^{2t}-1)^{3/2}}\exp\left({\frac{{-\theta}^2}{2(e^{2t}-1)}}\right) I_{[0,\infty]}(t) $$

where $t > 0$ and $\theta > 0$.

1) Show that the pdf of $T$ is a one parameter exponential family.

2) Find the pdf of $Y = \frac{1}{(e^{2T} - 1)}$.

3) Verify that the pdf of $Y = 1/(e^{2t} - 1)$ integrates to $1$.

4) Find $E[Y]$.

I am pretty sure I have 1 and 2. For 1 with the exponential family, I have:

$$ g(\theta) = \sqrt{\frac{2}{\pi}}\theta,\qquad h(t) = \frac{e^{2t}}{(e^{2t}-1)^{3/2}} I_{[0,\infty]}(t), $$

and $$ \eta(\theta) = {\theta}^2,\qquad T(t) = \frac{-1}{2(e^{2t}-1)}. $$ For this my variables are: $g(\theta)$ for my function that depends on $theta$, $h(t)$ for my function that depends on t, $\eta(\theta)$ that is my exponential for $/theta$, and $T(t)$ for my exponential for t.

For 2, I take my given in the value of $Y = \frac{1}{(e^{2T} - 1)}$ to get that $t = \frac{ln(\frac{1}{y}+1)}{2}$. Then I use these values and perform a transformation:

$ f(y\mid\theta) = \sqrt{\frac{2}{\pi}} \theta y^{3/2} e^{2t} \exp\left({\frac{{- y\theta}^2}{2}}\right)$ $|\frac{d}{dy}\frac{ln(\frac{1}{y}+1)}{2}|$

$= \sqrt{\frac{2}{\pi}} \theta y^{3/2} (\frac{1}{y} + 1) \exp\left({\frac{{- y\theta}^2}{2}}\right)$ $|\frac{d}{dy}\frac{ln(\frac{1}{y}+1)}{2}|$

$= \biggl \lbrace{\sqrt{\frac{2}{\pi}} \theta y^{1/2} \exp\left({\frac{{- y\theta}^2}{2}}\right) + \sqrt{\frac{2}{\pi}} \theta y^{3/2}\exp\left({\frac{{- y\theta}^2}{2}}\right)} \biggr \rbrace$ $|\frac{d}{dy}\frac{ln(\frac{1}{y}+1)}{2}|$

where $y > 0$ and $\theta > 0$. I am still not sure if this is exactly correct however. Am I missing something?

For numbers 3 and 4 I keep getting stuck trying to show it integrates to 1 so I am wondering if I transformed it incorrectly. Any assistance is greatly appreciated.

share|improve this question
You should really state your definition of an exponential family, since the notation of $c,h,w$ and $t$ could differ from others. –  Stefan Hansen Nov 19 '12 at 7:25
oour formula for $f(y\mid\theta)$ is incorrect since some of its parts still involve $t$. Additionally, I wonder whether you did not simply substitute $y$ for $t$ in the conditional density of $T$, forgetting the Jacobian. –  Did Nov 19 '12 at 7:55

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