Lognormal Probabilities and Quantiles

From Lognormal Questions to Normal Calculations

A lognormal model is useful when a quantity is positive and its logarithm is modeled by a normal distribution. For example, a hypothetical claim severity or service charge might be modeled this way. Probability questions are often asked in dollars or other original units, but the calculation is performed on the logarithmic scale.

The key operation is to take the natural logarithm of the threshold or interval endpoints. Since the natural logarithm is increasing, inequalities keep their direction. The resulting logarithms can be standardized and evaluated using the standard normal cumulative distribution function.

Definition: A random variable $X$ has a lognormal distribution with parameters $\mu$ and $\sigma^2$ if $X>0$ and $$ \ln X\sim N(\mu,\sigma^2), $$ where $\sigma>0$. Here $\mu$ and $\sigma^2$ are the mean and variance of $\ln X$, not of $X$. Write $\Phi(z)=P(Z\le z)$ for $Z\sim N(0,1)$.

Threshold Probabilities

For a positive threshold $x$, the event $X\le x$ is equivalent to $\ln X\le\ln x$. Standardizing the logarithm gives the basic probability formula. The same method applies to lower-tail, upper-tail, and interval probabilities.

P(X\le x) =P(\ln X\le\ln x) =\Phi\left(\frac{\ln x-\mu}{\sigma}\right),\qquad x>0.

The corresponding upper-tail probability is one minus this lower-tail probability. Because a lognormal variable is continuous, whether an endpoint is included does not change the probability. If $x\le 0$, then $P(X\le x)=0$, since $X$ is strictly positive; the logarithm formula is not used at a nonpositive threshold.

P(X>x) =1-\Phi\left(\frac{\ln x-\mu}{\sigma}\right),\qquad x>0.

Worked Example: Probability Above a Claim Threshold

For a hypothetical insurance portfolio, suppose an individual claim amount $X$, in dollars, satisfies $\ln X\sim N(6,0.4^2)$. Find the probability that a claim exceeds $e^{6.4}$ dollars. Also report the threshold approximately in dollars.

Method. Take the logarithm of the threshold, standardize on the log scale, and use a standard normal upper tail. The model parameters are $\mu=6$ and $\sigma=0.4$; the variance on the log scale is $0.16$.

The threshold is $x=e^{6.4}$ dollars, which is approximately $\$601.85$. Its standardized log value is $$ z=\frac{\ln(e^{6.4})-6}{0.4} =\frac{6.4-6}{0.4} =\frac{0.4}{0.4} =1. $$

Therefore, $$ P(X>e^{6.4})=1-\Phi(1) =1-0.8413447 =0.1586553 \approx 0.1587. $$

The probability is approximately $0.1587$, or $15.87\%$. As a check, the event is an upper tail above a threshold whose log-scale value is one standard deviation above $\mu$, so a probability below one half is consistent with the model.

Interval Probabilities

To find the probability that $X$ lies between two positive values, transform both endpoints. The standardized lower and upper endpoints need not be symmetric around zero. Subtract the lower standard normal cumulative probability from the upper one, using adequate precision before rounding the final result.

P(a<X<b) =\Phi\left(\frac{\ln b-\mu}{\sigma}\right) -\Phi\left(\frac{\ln a-\mu}{\sigma}\right), \qquad 0<a<b.

Why the interval formula holds: Since $\ln$ is strictly increasing, $a<X<b$ holds exactly when $\ln a<\ln X<\ln b$. Subtracting $\mu$ and dividing by the positive $\sigma$ preserves the inequalities, giving $$ \frac{\ln a-\mu}{\sigma} < \frac{\ln X-\mu}{\sigma} < \frac{\ln b-\mu}{\sigma}. $$ The middle quantity has the standard normal distribution. The probability between its two endpoints is the upper cumulative probability minus the lower cumulative probability, which gives the formula.

Worked Example: Probability Between Two Log-Scale Thresholds

A hypothetical service charge $X$, measured in dollars, has $\ln X\sim N(2,0.2^2)$. Find $P(e^{1.6}<X<e^{2.2})$. The stated endpoints are approximately $\$4.95$ and $\$9.03$.

Method. Transform each endpoint by taking its natural logarithm, then standardize. Here $\mu=2$, $\sigma=0.2$, and the requested event is an interval, not an upper tail.

For the lower endpoint, $$ z_{\mathrm{low}}=\frac{\ln(e^{1.6})-2}{0.2} =\frac{1.6-2}{0.2} =\frac{-0.4}{0.2} =-2. $$ For the upper endpoint, $$ z_{\mathrm{high}}=\frac{\ln(e^{2.2})-2}{0.2} =\frac{2.2-2}{0.2} =\frac{0.2}{0.2} =1. $$

Thus, $$ \begin{aligned} P(e^{1.6}<X<e^{2.2}) &=\Phi(1)-\Phi(-2)\\ &\approx 0.8413447-0.0227501\\ &=0.8185946\\ &\approx 0.8186. \end{aligned} $$

The probability is $0.8186$ to four decimal places. The subtraction uses unrounded cumulative probabilities before rounding the final answer; subtracting separately rounded table entries can give a different fourth decimal.

Lognormal Percentiles and Quantiles

A percentile reverses a probability calculation. The $p$th quantile $x_p$ is the value for which $P(X\le x_p)=p$, where $0<p<1$. Define $z_p$ as the standard normal $p$th quantile, so $\Phi(z_p)=p$. Taking logarithms turns the quantile condition into a normal quantile equation, which can then be solved for $x_p$.

Theorem: If $\ln X\sim N(\mu,\sigma^2)$, then the $p$th quantile of $X$ is $$ x_p=\exp(\mu+\sigma z_p),\qquad \Phi(z_p)=p,\quad 0<p<1. $$ The result is on the original scale of $X$, so its units are the same as those of $X$.

Identify the log-scale model.
Read $\mu$ and $\sigma$ from $\ln X\sim N(\mu,\sigma^2)$; do not treat $\sigma^2$ as the standard deviation.

Find the normal quantile.
Use a normal table or calculator to obtain $z_p$ such that $\Phi(z_p)=p$.

Return to the original scale.
Calculate $\exp(\mu+\sigma z_p)$, then round the resulting measurement as requested.

Worked Example: The 95th Percentile of a Cost

A hypothetical repair cost $X$, in dollars, has $\ln X\sim N(\ln 100,0.2^2)$. Find the 95th percentile, the amount that 95% of modeled costs do not exceed.

Method. This is an inverse probability calculation. The log-scale mean is $\mu=\ln 100$, the log-scale standard deviation is $\sigma=0.2$, and the desired normal quantile is $z_{0.95}\approx1.6448536$.

First calculate the log of the required percentile: $$ \ln x_{0.95} =\ln 100+(0.2)(1.6448536) =\ln 100+0.32897072. $$ Exponentiating both sides gives $$ \begin{aligned} x_{0.95} &=\exp(\ln 100+0.32897072)\\ &=100\exp(0.32897072)\\ &\approx100(1.3895376)\\ &\approx \$138.95376. \end{aligned} $$

Rounded to the nearest cent, the 95th percentile is $\$138.95$. As a check, its logarithm standardizes to approximately $1.6449$, whose standard normal cumulative probability is $0.95$.

Common Mistake: Standardizing the Original Variable

A lognormal variable is not normal on its original scale. Therefore, the expression $(x-\mu)/\sigma$ is not the standardization for $X$ when the model specifies $\ln X\sim N(\mu,\sigma^2)$. The normal mean and standard deviation apply to $\ln X$, so the threshold must be logged first.

For the claim example above, an incorrect calculation might be $$ \frac{e^{6.4}-6}{0.4} \approx\frac{601.85-6}{0.4} \approx1489.63, $$ followed by treating this as a standard normal value. This is wrong: it subtracts a log-scale parameter from a dollar amount and does not standardize a normal variable. The correct calculation is $$ \frac{\ln(e^{6.4})-6}{0.4}=1, $$ which gives the upper-tail probability $1-\Phi(1)\approx0.1587$.

Another frequent error in percentile problems is to stop at $\mu+\sigma z_p$. That quantity is the log of the percentile, not the percentile itself. Exponentiate to return to the units of $X$.

Key takeaway: For lognormal probabilities, transform positive thresholds with $\ln$, standardize using the mean and standard deviation of $\ln X$, and evaluate a normal probability. For quantiles, find the normal quantile first and exponentiate the resulting log-scale value.

Check Your Understanding

For each problem, use the stated lognormal model and round probabilities to four decimal places and dollar amounts to the nearest cent where applicable.

A hypothetical asset value $X$ has $\ln X\sim N(4,0.5^2)$. Find $P(X\le e^{4.5})$. (Answer: 0.8413)
A hypothetical payment amount $X$ has $\ln X\sim N(\ln 300,0.2^2)$. Find the probability that $X$ exceeds $300e^{0.4}$ dollars. (Answer: 0.0228)
A hypothetical cost $X$, in dollars, has $\ln X\sim N(3,0.4^2)$. Find its 75th percentile, using $z_{0.75}\approx0.67449$. (Answer: $26.31)
A hypothetical loss amount $X$ has $\ln X\sim N(4.8,0.3^2)$. Find $P(e^{4.5}<X<e^{5.25})$. (Answer: 0.7745)
A hypothetical account balance $X$, in dollars, has $\ln X\sim N(\ln 50,0.15^2)$. Find its 90th percentile, using $z_{0.90}\approx1.28155$. (Answer: $60.60)

What You'll Learn