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$$f(x;\theta) = 2x\theta\exp({-x^2})\left( \frac{\exp({-x^2})}{1-\exp({-x^2})}\right)^{\theta\ - 1}\mathbb I_{(\mathbb R_{++})}(x) $$ with $\theta \in \mathbb R_{++} $

does $f(x;\theta)$ belong to class exponentail of 1 parameter with $\theta$ unknown?

I have to write $f(x;θ) = a(θ)b(x)\exp(c(θ)d(x))$, I don't think this can be done. All I can do is write

$$2\theta \exp(lnx-x^2)\exp[(1-\theta).\log(1-e^{-x^2})] $$

Is wrong or right?

Type error.. I forgot a X in density. I already added it.. and b(x) ≥ 0 and a(\theta) ≥ 0

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  • $\begingroup$ Take the logarithm of $f$ (check your work carefully!) and see whether that splits into a function purely of $x,$ another function purely of $\theta,$ and a third function that is a product of a function of $\theta$ and a function of $x.$ Hint: that last function of $x$ may be complicated--it doesn't have to have a simple form. $\endgroup$ – whuber Feb 7 at 15:43
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\begin{align*} f\left(x;\theta\right) &= 2\cdot\theta\cdot\exp\left(-x^2\right)\cdot\left(\frac{\exp\left(-x^2\right)}{1-\exp\left(-x^2\right)}\right)^{\theta-1}\\ &= \underbrace{\theta}_{a\left(\theta\right)}\cdot\underbrace{2\cdot\exp\left(-x^2\right)}_{b\left(x\right)}\cdot\exp\left(\underbrace{\theta-1}_{c\left(\theta\right)}\cdot\underbrace{\log\left(\frac{\exp\left(-x^2\right)}{1-\exp\left(-x^2\right)}\right)}_{d\left(x\right)}\right)\\ \end{align*}

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  • $\begingroup$ Thank & and sorry, I forgot a X in the density $\endgroup$ – Chedard Feb 7 at 5:27

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