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My idea is that the Expectation of Cauchy Distribution is infinite, and I find the following lemma

If $X$ were from the exponential family, it would have finite expectation.

The question is solved if I can prove the lemma, but I have no idea how to prove it.

I could get that if $f_X(x;θ)$ can be written as $$f_X(x;\eta)=h(x)\exp\left(\sum_{i=1}^s\eta_iT_i(x)−A(\eta)\right)$$ then $E[T_i(X)]=\frac{\partial A(\eta)}{\partial η_i}$ is finite

But I don't know how to apply this to show that the expectation $E[X]$ is finite.

G Frazao
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MartinS
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    Have you tried showing directly that the density of the Cauchy distribution cannot be written in the exponential form? Your idea should also work fine, but I'm not sure it is easier than the straightforward way. – G Frazao Sep 29 '22 at 10:28
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    @GFrazao I have tried so, but I think it's much harder to show strictly that the density CAN'T be written in a form than comfirm that it CAN, so I turn to another way. If you have any other method, please ignore my idea and show my how to solve it. – MartinS Sep 29 '22 at 14:17

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Distributions of the Exponential Family can be written in the form: $$ f_X(x; \theta) = h(x)g(\theta)\exp\biggl(\eta(\theta)\cdot T(x)\biggr)$$

The Cauchy distribution has the following expression: $$\begin{align} f_X(x; \lambda) &= \frac{\lambda}{\pi(\lambda^2+x^2)} \\ &= \frac{\lambda}{\pi}\exp\biggl(-\log (\lambda^2+x^2) \biggr) \tag{*}\label{*}\\ &= \frac{\lambda}{\pi}\exp\biggl(-2\log (\lambda) \log \bigl(1+\frac{x^2}{\lambda^2}\bigr)\biggr) \tag{*} \end{align}$$

Can you fill in the final step?

G Frazao
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