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Given $X$ ~ $LogN(\mu, \sigma^2)$, I want to prove that the distribution function of $X$ is given by $\Phi\Big(\frac{\ln(x)-\mu}{\sigma}\Big)$, where $\Phi()$ is the distribution function of a $N(0,1)$-distributed r.v.

I know that the density of $X$ is given by

$$ f(x) = \frac{1}{\sqrt{2\pi}\sigma x}\exp\{-\frac{1}{2} \Big(\frac{\ln(x)-\mu}{2\sigma}\Big)^2\}$$

so we get

$$ F(x) = \frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^x\exp\{-\frac{1}{2} \Big(\frac{\ln(t)-\mu}{2\sigma}\Big)^2\}\frac{1}{t}dt$$ $$= \frac{1}{\sqrt{2\pi}}\int_{0}^{\frac{\ln(x)-\mu}{\sigma}}\exp\{-\frac{1}{2} z^2\}dz = \Phi\Big(\frac{\ln(x)-\mu}{\sigma}\Big)$$

Is it correct, that the lower bound transforms to $0$?

Furthermore, how can one estimate the parameters $(\mu, \sigma^2)$ for given data using the Maximum-Likelihood-method for the r.v. $\ln(X)$?

For the second question I think we can deduce this to the ML-estimators of the normal distribution, since

$$ L(x; \mu, \sigma^2) = \prod \Big(\frac{1}{x_i}\Big) N(\ln(x); \mu, \sigma^2) $$

where $L$ is the density of the log-normal distribution and $N$ is the density of the normal distribution. For the log-lieklihood this means

$$ l_L(x; \mu, \sigma^2) = -\sum_k \ln(x_k) +l_N(\ln(x); \mu, \sigma^2) $$

If we differentiate this, the sum of the logarithms always cancels, since it is constant, and therefore it is enough to know the ML-estimators for the normal distribution. Therefore we get in the end that

$$ \hat{\mu}=\frac{1}{n}\sum_k \ln(x_k), \text{ } \hat{\sigma}^2=\frac{1}{n}\sum_k (\ln(x_k)-\hat{\mu})^2.$$

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Given $X \sim \operatorname{LogNormal}(\mu, \sigma^2)$, then by definition, $X = e^Y$ for $Y \sim \operatorname{Normal}(\mu,\sigma^2)$, hence $X = e^{\sigma Z + \mu}$ for $Z \sim \operatorname{Normal}(0,1)$. It immediately follows that $$F_X(x) = \Pr[X \le x] = \Pr[e^{\sigma Z + \mu} \le x] = \Pr\left[Z \le \frac{\log x - \mu}{\sigma}\right] = \Phi\left(\frac{\log x - \mu}{\sigma}\right).$$ It also follows from the above that the density of $X$ is $$f_X(x) = F'_X(x) = \frac{d}{dx} \left[\Phi\left(\frac{\log x - \mu}{\sigma}\right)\right] = \frac{1}{\sigma x} f_Z\left(\frac{\log x - \mu}{\sigma}\right) = \frac{1}{\sqrt{2\pi} \sigma x} e^{-(\log x - \mu)^2/(2\sigma^2)}.$$

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  • $\begingroup$ True, that would be the short way without specifiying the integrals. Thank you! $\endgroup$ – ducks17 Mar 28 '17 at 17:27
  • $\begingroup$ And what about the secon question? $\endgroup$ – ducks17 Mar 28 '17 at 19:18
  • $\begingroup$ @ducks17 Show what you have tried or what you know about maximum likelihood estimation, and I may address that part accordingly. As it stands, that part of your question is too broad for me to furnish a succinct answer. $\endgroup$ – heropup Mar 28 '17 at 19:20
  • $\begingroup$ See the post for my way of solving the problem. Do you think it's correct? $\endgroup$ – ducks17 Mar 28 '17 at 19:51
  • $\begingroup$ @ducks17 Yes, it is correct. You can from the very beginning note that $\ln(X_i)$ have Normal distribution and take likelihood function for it. $\endgroup$ – NCh Mar 29 '17 at 1:29

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