Lognormal Distribution

Practice Problem Set 3 – basic lognormal problems

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This post has several practice problems to go with this previous discussion on lognormal distribution.

 Practice Problem 3A The amount of annual losses from an insured follows a lognormal distribution with parameters $\mu$ and $\sigma$ = 0.6 and with mode = 2.5. Calculate the mean annual loss for a randomly selected insured.
 Practice Problem 3B Claim size for an auto insurance coverage follows a lognormal distribution with mean 149.157 and variance 223.5945. Determine the probability that a randomly selected claim will be greater than 170.
 Practice Problem 3C For x-ray machines produced by a certain manufacturer, the following is known. Lifetime in years follows a lognormal distribution with $\mu$ = 0.9 and $\sigma$. The expected lifetime of such machines is 15 years. Calculate the probability that an x-ray machine produced by this manufacturer will last at least 12 years.
 Practice Problem 3D Claim sizes expressed in US dollars follow a lognormal distribution with parameters $\mu$ = 5 and $\sigma$ = 0.25. One Canadian dollar is currently worth \$0.75 US dollars. Calculate the 75th percentile of a claim in Canadian dollars.
 Practice Problem 3E For a commercial fire coverage, the size of a loss follows a lognormal distribution with parameters $\mu$ = 2.75 and $\sigma$ = 0.75. Determine $y-x$ where $y$ is the 75th percentile of a loss and $x$ is the 25th percentile of a loss. Note that $y-x$ is known as the interquartile range.
 Practice Problem 3F Claim sizes in the current year follow a lognormal distribution with $\mu$ = 4.75 and $\sigma$ = 0.25. In the next year, all claims are expected to be inflated uniformly by 25%. One claim is expected in the next year for an insured. Determine $y-x$ where $y$ is the 80th percentile of the size of this claim and $x$ is the 40th percentile of the size of this claim.
 Practice Problem 3G In the current year, losses follow a lognormal distribution with $\mu$ = 1.6 and $\sigma$ = 1.35. In the next year, inflation of 20% will impact all losses uniformly. Determine the median of the portion of next year’s loss distribution that is above 10.
 Practice Problem 3H Losses follow a lognormal distribution with mean 17 and variance 219. Determine the skewness of the loss distribution.
 Practice Problem 3I Losses from an insurance coverage follow a lognormal distribution with parameters $\mu$ and $\sigma$ = 2. The 80th percentile of the losses is 5884. Determine the probability that a loss is less than 5000.
 Practice Problem 3J Losses from an insurance coverage follow a lognormal distribution. The 25th percentile of the losses is 133.62. The 75th percentile of the losses is 997.25. Determine the mean of the losses.

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

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3A 4.29
3B 0.0869
3C 0.2033
3D 233.9675
3E 16.39085
3F 42.5155
3G 21.143268
3H 3.271185
3I 0.7764
3J 1124.394559

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$\copyright$ 2017 – Dan Ma

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.

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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 $k$th 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 $\displaystyle e^{\mu - \sigma^2}$

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Examples

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 $P(Y_1 > 0.9)$

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

\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}

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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)

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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 $Y$.

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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}

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$\copyright$ 2017 – Dan Ma