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(a) Find CDF for $Y$.

(b) Then use the CDF to show that $Y$ has the following PDF

$$f(y)=\frac{1}{y \sqrt{2 \pi \sigma^{2}}} \exp \left(-\frac{(\log y-\mu)^{2}}{2 \sigma^{2}}\right), \quad y>0$$

My problem is that I'm not allowing to use change of variables! I have to do this without.

I'm not sure how to start. A gentle push forward will be much appreciated.

Sorry
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  • Take $y > 0$. Then \begin{equation} \mathbb{P}(Y \leq y) = \mathbb{P}(X \leq \log y) \end{equation} because $\log$ is increasing on its domain. For part (b) think about the relationship between cdf's and pdf's. – pg_star Mar 03 '20 at 21:40
  • When I differentiate CDF i get PDF. The hard part is actually finding the CDF. – Sorry Mar 03 '20 at 21:55

2 Answers2

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Hint

$$P(Y\leq a)=P(e^X\leq a)=P(X\leq \log(a))=\Psi(\log(a))$$ $$=\int_{-\infty}^{\log(a)} \frac{1}{\sqrt{2\pi \sigma^2}} e^{\frac{-(t-\mu)^2}{2\sigma^2}} dt$$ Now derive by $a$. Leibniz_integral_rule

$$f_x(a)=\frac{d}{da} F_X(a)=\frac{d}{da} \int_{-\infty}^{\log(a)} \frac{1}{\sqrt{2\pi \sigma^2}} e^{\frac{-(t-\mu)^2}{2\sigma^2}} dt$$

$$= (\frac{d}{da}\log(a)) \frac{1}{\sqrt{2\pi \sigma^2}} e^{\frac{-(\log(a)-\mu)^2}{2\sigma^2}} $$

$$= \frac{1}{a} \frac{1}{\sqrt{2\pi \sigma^2}} e^{\frac{-(\log(a)-\mu)^2}{2\sigma^2}} $$

Masoud
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To get CDF for $y\ge 0$: $P(Y\le y)=P(e^X\le y)=P(X\le lny)=\frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{lny}e^{-\frac{(w-\mu)^2}{2\sigma^2}}dw$.

To get the PDF simply take the derivative to get: PDF $=\frac{1}{y\sqrt{2\pi \sigma^2}}e^{-\frac{(lny-\mu)^2}{2\sigma^2}}$

herb steinberg
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