Basic properties of lognormal distribution

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A detailed discussion of the mathematical properties of lognormal distribution is found in this previous post in a companion blog. This post shows how to work basic calculation problems for lognormal distribution. A summary of lognormal distribution is given and is followed by several examples. Practice problems are in the next post.


Basic Properties

The random variable Y is said to follow a lognormal distribution with parameters \mu and \sigma if \log(Y) follows a normal distribution with mean \mu and variance \sigma^2. Here, \log is the natural logarithm in base e = 2.718281828…. It is difficult (if not impossible) to calculate probabilities by integrating the lognormal density function. Since the lognormal distribution is intimately related to the normal distribution, the basic lognormal calculation is performed by calculating the corresponding normal distribution. The following summary shows how.

In the following points, Y has a lognormal distribution with parameters \mu and \sigma and X=\log(Y) is the corresponding normal distribution with mean \mu and variance \sigma^2. The notation \exp(t) means raising e to the number t.

1. Lognormal observations and normal observations
  • Taking natural log of a lognormal observation gives a normal observation.
  • Raising e to a normal observation gives a lognormal observation.
2. Lognormal CDF and normal CDF
  • \Phi(z) is the CDF of the standard normal distribution.
  • \displaystyle F_Y(y)=\Phi \biggl(\frac{\log(y)-\mu}{\sigma} \biggr)
  • In words, lognormal CDF evaluated at y equals to standard normal CDF evaluated at \frac{\log(y)-\mu}{\sigma}.
    • Derivation:
      \displaystyle \begin{aligned} F_Y(y)=P(Y \le y)&=P[\log(Y) \le \log(y)] \\&=P \biggl[\frac{\log(Y)-\mu}{\sigma} \le \frac{\log(y)-\mu}{\sigma} \biggr] \\&=\Phi \biggl(\frac{\log(y)-\mu}{\sigma} \biggr)  \end{aligned}

3. Lognormal density function and normal density function
  • normal density: \displaystyle f_X(x)=\frac{1}{\sqrt{2 \pi} \ \sigma} \exp(-\frac{(x-\mu)^2}{2 \sigma^2})
  • lognormal density: \displaystyle F_Y(y)=\frac{1}{\sqrt{2 \pi} \ \sigma \ y} \exp(-\frac{(\log(y)-\mu)^2}{2 \sigma^2})
  • In words, lognormal density evaluated at y equals to normal density evaluated at \log(y) times \frac{1}{y}.
4. Lognormal moments and normal moment generating function
  • normal mgf: M(t)=e^{\mu \ t+\frac{1}{2} \sigma^2 \ t^2}=\exp(\mu \ t+\frac{1}{2} \sigma^2 \ t^2)
  • lognormal moment: E(x^k)=M(k)=\exp(\mu \ k+\frac{1}{2} \sigma^2 \ k^2)
  • In words, lognormal kth raw moment equals to normal mgf evaluated at k.
5. Examples of lognormal moments
  • E(Y)=E[e^{\log(Y)}]=M(1)=e^{\mu+\frac{1}{2} \sigma^2}
  • E(Y^2)=E[e^{2 \log(Y)}]=M(2)=e^{2 \mu+2 \ \sigma^2}
  • Var(Y)=e^{2 \ \mu+2 \ \sigma^2}-e^{2 \mu+ \sigma^2}=e^{2 \ \mu+\sigma^2} (e^{\sigma^2}-1)
  • skewness: \gamma_1=(e^{\sigma^2}+2) \sqrt{e^{\sigma^2}-1}
  • kurtosis: e^{4 \sigma^2}+2 e^{3 \sigma^2}+3 e^{2 \sigma^2}-3
6. Lognormal percentiles and normal percentiles
  • (100p)th percentile of the normal distribution is \displaystyle z_p.
  • (100p)th percentile of the normal distribution with mean \mu and variance \sigma^2 is \displaystyle \mu+z_p \times \sigma.
  • (100p)th percentile of the lognormal distribution with parameters \mu and \sigma is \displaystyle e^{\mu+z_p \times \sigma}.
  • In words, to find the (100p)th percentile of the lognormal distribution, find the (100p)th percentile of the corresponding normal distribution and then raise e to it.
7. Constant multiple of lognormal distribution
  • Let c>0. If Y has a lognormal distribution with parameters \mu and \sigma, then c Y has a lognormal distribution with parameters \mu+\log(c) and \sigma.
  • The effect of the multiplicative constant is on the parameter \mu in the form of an additive adjustment of \log(c).
8. Mode of lognormal distribution
  • \displaystyle e^{\mu - \sigma^2}



Two examples are given to illustrate the calculation discussed here. The next post has practice problems.

All normal probabilities are obtained by using the normal distribution table found here.

Example 1
Suppose that the random variable Y has a lognormal distribution with parameters \mu = 1 and \sigma = 2. Calculate the following.

  • P(Y \le 75.19) and P(Y > 0.9)
  • The 67th, 95th and 99th percentiles of Y.
  • Let Y_1=1.1Y. Find P(Y_1 \le 75.19) and P(Y_1 > 0.9)

\text{ }

    \displaystyle P(Y \le 75.19)=\Phi \biggl(\frac{\log(75.19)-1}{2} \biggr)=\Phi(1.66)=0.9515

    \displaystyle \begin{aligned}P(Y > 0.9)&=1-\Phi \biggl(\frac{\log(0.9)-1}{2} \biggr) \\&=1-\Phi(-0.55) \\&=1-(1-0.7088) \\&=0.7088  \end{aligned}


To find the percentiles, first find the standard normal percentiles, either by using calculator or by looking up a table. Using a standard normal table, we get 0.44 (67th percentile), 1.645 (95th percentile) and 2.33 (99th percentile). The following gives the lognormal percentiles.

    \displaystyle e^{1+0.44(2)}=e^{1.88} = 6.5535 (67th percentile)

    \displaystyle e^{1+1.645 (2)}=e^{4.29} = 72.9665 (95th percentile)

    \displaystyle e^{1+2.33(2)}=e^{5.66} = 287.1486 (99th percentile)


The random variable Y_1=1.1Y has a lognormal distribution with parameters \mu=1+\log(1.1) and \sigma = 2.

    \displaystyle P(Y_1 \le 75.19)=\Phi \biggl(\frac{\log(75.19)-1-\log(1.1)}{2} \biggr)=\Phi(1.61)=0.9463

    \displaystyle \begin{aligned}P(Y_1 > 0.9)&=1-\Phi \biggl(\frac{\log(0.9)-1-\log(1.1)}{2} \biggr) \\&=1-\Phi(-0.57) \\&=1-(1-0.7157) \\&=0.7157  \end{aligned}

Note. One interpretation of Y_1=1.1Y is that of inflation, in this case a 10% inflation. For example, let Y be the size of a randomly selected auto insurance collision claim in the current year. If the claims are expected to increase 10% in the following year, Y_1=1.1Y is the the size of a randomly selected claim in the following year.

Example 2
Suppose that the random variable Y has a lognormal distribution with mean 12.18 and variance 255.02. Calculation the following.

  • P(Y > 10)
  • The skewness and kurtosis of Y.

\text{ }

First, determine the parameters \mu and \sigma by setting up the following equations.

    \displaystyle E(Y)=e^{\mu+\frac{1}{2} \sigma^2}=12.18

    \displaystyle Var(Y)=[e^{\sigma^2}-1] \ [  e^{ \mu+\frac{1}{2} \sigma^2} ]^2=255.02

Plug the first equation into the second equation and obtain the equation \displaystyle [e^{\sigma^2}-1] \ [12.18 ]^2=255.02. Solving for \sigma produces \sigma = 1. Plug \sigma = 1 into the first equation produces \mu = 2. The following gives the desired probability.

    \displaystyle P(Y > 10)=1-\Phi \biggl(\frac{\log(10)-2}{1} \biggr)=1-\Phi(0.30)=1-0.6179=0.3821

To find the skewness and kurtosis, one way is to find the first 4 lognormal moments and then calculate the third standardized moment (skewness) and the fourth standardized moment (kurtosis). To see how this is done, see this previous post. Another is to use the formulas given above.

    \gamma_1=(e^{\sigma^2}+2) \sqrt{e^{\sigma^2}-1}=(e^1+2) \sqrt{e^1-1}=6.1849

    \text{Kurtosis}=e^{4 \sigma^2}+2 e^{3 \sigma^2}+3 e^{2 \sigma^2}-3=e^{4}+2 e^{3}+3 e^{2}-3=113.9364

Example 3
Suppose that the lifetime (in years) of a certain type of machines follows the lognormal distribution described in Example 2. Suppose that you purchased such a machine that is 10-year old. What is the probability that it will last another 10 years?

This is a conditional probability since the machine already survived for 10 years already.

    \displaystyle \begin{aligned}P(Y > 20 \lvert Y > 10)&=\frac{1-\Phi (\frac{\log(20)-2}{1} )}{1-\Phi (\frac{\log(10)-2}{1} )} \\&=\frac{1-\Phi(1.0)}{1-\Phi(0.30)} \\&=\frac{1-0.8413}{1-0.6179} \\&=\frac{0.1587}{0.3821}=0.4153  \end{aligned}

\copyright 2017 – Dan Ma


5 thoughts on “Basic properties of lognormal distribution

    […] post has several practice problems to go with this previous discussion on lognormal […]

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