Let $X$ be $N (\mu,\sigma^2)$.Define the random variable $Y=e^x$ and find its probability distribution function. My solution is this, let $G(y)= P(Y\le y)=P(e^x\le y) =P(X\le ln y)$.Let $f(x)=\frac{1}{\sigma\sqrt{2\pi}} \exp((x-\mu)^2/2\sigma^2))$.Thus,$$P(X\le ln y)=\int_{-\infty}^{\ln y} f(x)\,dx$$Please help me with the chain rule. How can find the pdf?Thank you
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Hint: You have $P(X\leq ln(y))$. This is $F_x(ln(y))$. $F_x(ln(y))$ is the cdf of normal distribution with $X\sim \mathcal N(\mu_x,\sigma_x^2)$
To find $g_Y(y)$ you have to differentiate $F_x(ln(y))$. Don´t forget using the chain rule.
callculus42
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Please help me with the chain rule. thanks – Norberto Vergara Sr. Nov 19 '14 at 14:43
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It is $g_Y(y)=\frac{d}{dy}F_x(ln(y))=f_x(ln(y))\cdot \frac{1}{y}$, because $\left( ln(y)\right)'=\frac{1}{y}$ – callculus42 Nov 19 '14 at 15:01
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This follows by the change-of-variables formula for densities: if $X$ has density $f_X$ and $g$ is one-to-one, then $Y=g(X)$ has density $$ f_Y(y)=\bigg|\frac{\mathrm d}{\mathrm d y}g^{-1}(y)\bigg| f_X(g^{-1}(y)). $$
Stefan Hansen
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