# How to find the area undernearth a log-normal curve

I wanted to find the area underneath a Gaussian distribution. I found online that for an equation of the form:

$Ne^{-\frac{(x-\mu)^2}{2\sigma^2}}$

The area under the curve is given by:

$N \sigma \sqrt{2 \pi}$

My question is:

I am now looking at a log-normal distribution:

$\frac{1}{x} Ne^{-\frac{(ln(x)-\mu)^2}{2\sigma^2}}$

I'm not sure how to find the area underneath the log-normal distribution curve.

Thank you.

• Use the substitution y=ln(x) (note that dy = 1/x dx) – Glen_b -Reinstate Monica Sep 3 '18 at 0:13
• Okay so I did the calculation, and things ended up cancelling so that I was left with my original equation but with y instead of x. If this is correct, then I believe the conclusion that I am drawing, is that the area under a normal distribution and a log-normal distribution are the same (providing they have the same mean and standard deviation values). Does this seem correct? – user1551817 Sep 3 '18 at 0:57
• The area under a lognormal density and under a normal density are actually both 1. What you're doing is computing the normalizing constants that should already be there (you can see the reciprocal of those normalizing constants in the densities given at both those links). Yes, they're the same, because a lognormal random variable is just a transformed normal random variable. You may wish to post your answer. – Glen_b -Reinstate Monica Sep 3 '18 at 1:01

As suggested by Glen_b, I just used the substitution:

$y = ln(x)$

and

$dy = \frac{1}{x}dx$, therefore: $dx = e^y dy$

When these substitutions are plugged into the integration (without any calculation needing being done), the above terms cancel and you are left with the original integration.

Also as pointed out by Glen_b, the area under the probability density of the normal distribution is defined as 1. Therefore, it is 1 for the lognormal distribution too.

So if I have a lognormal distribution of the form:

$\frac{1}{x}Ne^{-\frac{(ln(x)-\mu)^2}{2\sigma^2}}$

The area under the curve will still be given by:

$N \sigma \sqrt{2 \pi}$