Let $X$ be a random variable with normal distribution with parameters: $$m = 1$$ and $$\sigma = 2$$ How can the probability density function of $$Z = -\frac{\ln |X|}{3}$$ be determined?
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asked Apr 24, 2015 at 11:29
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First of all, note that if $\mathbf{P}_Z$ is the law of $Z$, then you can check that $\mathbf{P}_Z$ is absolutely continuous with respect to $\lambda$ (the Lebesgue measure), so there will be a density function. Now to find it, use the cumulative distribution function. Let $t \in \mathbb{R}$.
$$F(t) = \mathbf{P}(Z \leq t) = \mathbf{P}(e^{-3t}\leq|X|)$$ This can be explicitly computed because you know the distribution of $X$ : $$F(t) = 1 - \int_{-e^{-3t}}^{e^{-3t}}\frac{e^{-(x-\mu)^2/2\sigma}}{\sqrt{2\pi}\sigma}dx$$
To find the density function, just take the derivative of $F$.
answered Apr 24, 2015 at 11:35
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$\begingroup$ @Herbet Quain The problem is I don't know how to solve this integral. $\endgroup$ Apr 24, 2015 at 11:43
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1$\begingroup$ @JohnG. You don't need to solve or compute the value of the integral as a function of $t$ in order to takes its derivative with respect to $t$. Since $t$ occurs only in the limits, just apply the Fundamental Theorem of Calculus to get the derivative directly. $\endgroup$ Apr 24, 2015 at 12:25
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$\begingroup$ @Did : thank you, I just edited my answer. $\endgroup$ Apr 24, 2015 at 12:29