1
$\begingroup$

Let's assume we have two normal distributions ($G_1$, $G_2$) with mean $\mu_1$, $\mu_2$ and standard deviation $\sigma_1$, $\sigma_2$, respectively. If we generate random numbers ($x_1$) following $G_1$ and we apply the transformation: $$ x_1 \rightarrow \frac{(x_1 - \mu_1 + \mu_2 \sigma_1/ \sigma_2)}{\sigma_1/ \sigma_2} $$ at the end the distribution of $x_1$ will follow $G_2$: mean $\mu_2$ and standard deviation $\sigma_2$.

If we want to do the same thing with two lognormal distributions with location parameter $m_1$, $m_2$ and scale parameter $s_1$, $s_2$, the transformation reads: $$ x_1 \rightarrow \exp \left({\frac{(\log{x_1} - m_1 + m_2 s_1/ s_2)}{s_1/ s_2}} \right)$$

What about the sum of normal and a lognormal distribution, such as: $$ G(x) = a \, \frac{1}{\sqrt{2\pi}\sigma} \exp \left( - (x-\mu)^2/\sigma^2\right) + b \, \frac{1}{\sqrt{2\pi}s x} \exp \left( - (\log(x)-m)^2/s^2\right)$$ is there an equivalent transformation to map one of such functions to another?

$\endgroup$
  • $\begingroup$ Do you mean the sum in the sense: sum of two random variables (whose density is the convolution of the densities) or do you mean the sum in the sene: a r.v. with density a weighted sum of the densities? The latter one seems to be the one you wrote down. $\endgroup$ – Kore-N Feb 12 '17 at 14:16
  • $\begingroup$ I mean the second one, indeed. $\endgroup$ – jacopo.margutti Feb 12 '17 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.