Respuesta :
Answer:
[tex] \frac{1}{X} \sim log N(-\mu , \sigma^2)[/tex]
Step-by-step explanation:
For this case we know that the distribution for the random variable is given by:
[tex]X \sim logN(\mu ,\sigma^2)[/tex]
The density function for the log normal random variable is given by:
[tex] f)x,\sigma) = \frac{1}{x \sigma \sqrt{2\pi}}e^{- \frac{ln x^2}{2\sigma^2}}[/tex]
And we want to find the distribution for the random variable [tex]\frac{1}{X}[/tex]
In order to find this distribution we can use the cumulative distribution function like this:
Let [tex] Y= \frac{1}{X}[/tex], if we solve for X from this transformation we got:
[tex] X= \frac{1}{Y}[/tex]
And then we have this:
[tex] F_Y (y) = P(Y \leq y) = P(X \leq \frac{1}{y}) = F_X (\frac{1}{y})[/tex]
And we can find the density function as the derivate of the distribution function like this:
[tex] f_Y (y) = F'_Y (y) = -\frac{1}{y^2} f_Y(\frac{1}{y})[/tex]
[tex] f_Y (y)= -\frac{1}{y^2} \frac{1}{\frac{1}{y} \sigma \sqrt{2\pi}} e^{- \frac{ln(\frac{1}{y})^2}{2\sigma^2}}[/tex]
But we see that we don't have an specified form for the distribution
If we assume that X follows a normal distribution [tex] X\sim N (\mu_z,\sigma^2_z)[/tex] and we use the transformation [tex]X=e^Y[/tex] we see that X follows a log normal distribution. And we see that:
[tex]\frac{1}{X}= \frac{1}{e^Y}=e^{-Y}[/tex]
And if we find the distribution of [tex]e^{-y}[/tex] we got this:
[tex] f_Y (y) = F'_Y (y) = -e^{-y} f_Y(e^{-y})[/tex]
[tex] f_Y (y)= -e^{-y} \frac{1}{e^{-y} \sigma \sqrt{2\pi}} e^{- \frac{ln(e^{-y})^2}{2\sigma^2}}[/tex]
[tex] f_Y (y) = -\frac{1}{\sigma \sqrt{2\pi}}e^{\frac{y^2}{2\sigma^2}}[/tex]
And then we see that [tex]Y= \frac{1}{X} \sim log N(-\mu , \sigma^2)[/tex]
