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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.

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  • $\begingroup$ This is really seems to be a question about notational conventions, of which there are several for the lognormal convention. However, one thing is clear: because the logarithm is an order-preserving, one-to-one way of re-expressing positive numerical values, the statement $\Pr(x\lt 0.90)=\Pr(\log(x)\lt\log(0.90))$ is always true, no matter what the probability distribution might be. $\endgroup$ – whuber Apr 4 '14 at 15:37
  • $\begingroup$ It's true because logs are monotonic. $\endgroup$ – Glen_b Apr 5 '14 at 6:43
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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.

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    $\begingroup$ You need more than that the logarithm is one-to-one: your conclusion requires the fact that the log is measurable and everywhere strictly increasing. $\endgroup$ – whuber Apr 4 '14 at 15:39
  • $\begingroup$ And when I mention measurable, you would say ... define the metastability! So I will leave it for others to provide a rigorous proof ;) $\endgroup$ – Stat Apr 4 '14 at 15:42
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    $\begingroup$ This is not an issue of rigor but simply of being correct. The strictly increasing requirement is essential for even understanding why your answer works. It is not at all technical or difficult to understand. It is needed for two reasons: first, one-to-one decreasing functions like $x\to 1/x$ (on the positive reals) reverse the directions of inequalities. Second, many one-to-one functions do not respect inequalities at all. Consider, as a counterexample, the function $x\to 1-x$ for $0\lt x\lt 1$ and $x\to x$ for $x\ge 1$. $\endgroup$ – whuber Apr 4 '14 at 15:46

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