If I have a variable x that is lognormal(mu=0, sd=.1), and say I want to compute P(x < .90)
Can I then say P(x < .90) = P( log(x) < log(.90) ) ? Going by the book I am reading from, this appears to be true, but I don't get why.
I know that log(x) is normal(mu=0, sd=.1), but I don't get why the cdf value corresponding to log(.90) from the normal distribution has the same cdf value as .90 from the lognormal distribution.
Best Answer
Let me give you an example where you have probably seen before. Suppose $X\sim N(\mu, \sigma^2)$ i.e. normally distributed with mean $\mu$ and variance $\sigma^2$. Then what is$P(X<a)$? To answer this question, since we have tables for standard normal distribution i.e. $N(0, 1)$, we standardize it as follows: $$P(X<a)=P(\dfrac{X-\mu}{\sigma}<\dfrac{a-\mu}{\sigma})=P(Z<\dfrac{a-\mu}{\sigma}),$$ where $Z\sim N(0,1)$ i.e. $Z$ is normally distributed with mean $0$ and variance $1$. The same argument applies here. Since the $\log$ is a 1-1 function that preserves the signs of in-equality, you can apply it to the both sides of the in-equality without changing the sign. But that helps you to find the probability since now the $\log(X)\sim N(0,1)$. So, to answer your question, yes you definitely can take log.