I just wanted to verify that my attempt is correct. Thank you in advance for reading this.

A distribution is said to belong in the exponential family if its probability density function can be written in the form $$f_Y(y;\theta)=\exp\Big(\sum_{j=1}^na_j(\theta)b_j(y)+c(\theta)+d(y)\Big).$$ Now, we have $$f_Y(y)=\frac{1}{\sqrt{2\pi\sigma^2}y}\exp\Big(-\frac{1}{2\sigma^2}(\log(y)-\mu)^2\Big).$$ Then, taking logarithms \begin{align*} \log(f_Y(y))&=\log\Big(\frac{1}{\sqrt{2\pi\sigma^2}y}\exp\Big(-\frac{1}{2\sigma^2}(\log(y)-\mu)^2\Big)\Big)\\ &=\log\Big((2\pi\sigma^2)^{-\frac{1}{2}}y^{-1}\exp\Big(-\frac{1}{2\sigma^2}(\log(y)-\mu)^2\Big)\Big)\\ &=-\frac{1}{2}\log(2\pi\sigma^2)-\log(y)-\frac{1}{2\sigma^2}(\log(y)^2-2\mu\log(y)+\mu^2))\\ &=-\frac{1}{2}\log(2\pi\sigma^2)-\log(y)-\frac{\log(y)^2}{2\sigma^2}+\frac{\mu}{\sigma^2}\log(y)-\frac{\mu^2}{2\sigma^2}. \end{align*} Now, exponentiating $$f_Y(y)=\exp\Bigg(-\frac{1}{2}\log(2\pi\sigma^2)-\log(y)-\frac{\log(y)^2}{2\sigma^2}+\frac{\mu}{\sigma^2}\log(y)-\frac{\mu^2}{2\sigma^2}\Bigg).$$ So, we identify $$a_1(\mu,\sigma^2)=-\frac{1}{2\sigma^2}, b_1(y)=\log(y)^2, a_2(\mu,\sigma^2)=\frac{\mu}{\sigma^2}, b_2(y)=\log(y),$$ $$c(\mu,\sigma^2)=\frac{1}{2}\log(2\pi\sigma^2)-\frac{\mu^2}{2\sigma^2}, d(y)=-\log(y).$$ Thus, this distribution belongs in the exponential family.

If this isn't correct - then where have I gone wrong?

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    $\begingroup$ Is this for homework? Consider reading the tag self-study and applying the tag if necessary. $\endgroup$ – Firebug Jun 3 '18 at 15:10
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    $\begingroup$ Why take logs and then exponentiate back to the original structure? You can just expand the square term and you'll get the same results. Note also that the term $c(\theta)$ doesn't have to be moved inside the exponential term, as $\exp{c(\theta)}$ can just be rewritten as, say, $c'(\theta)$ without the exponential. $\endgroup$ – jbowman Jun 3 '18 at 15:16
  • $\begingroup$ I see what you mean - I guess it didn't occur to me that it isn't necessary to move everything inside the exponential all the time. Thank you for the pointers. $\endgroup$ – thesmallprint Jun 3 '18 at 15:25
  • $\begingroup$ If you have already shown that the Normal distribution belongs to an exponential family, you do not need to do it again for the log-Normal. $\endgroup$ – Xi'an Jun 3 '18 at 15:34

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