Question

Individual losses on an insurance portofolio, X, are modelled as a lognormal random variable with parameter μ=2 and σ=0.5. Let S be the aggregate (total) loss for 100 independent individual losses. The distribution of S can be approximated by that of a normal distribution with appropriate parameters. Calculate the probability that an individual loss exceeds 10

Answer 1

To find the probability of an individual loss exceeding 10 in a lognormal distribution with parameters μ=2 and σ=0.5, we calculate the z-score and then look it up in the standard normal distribution table to find the desired probability.

Step-by-step explanation:

The student wants to calculate the probability that an individual loss exceeds 10 when losses follow a lognormal distribution with parameters μ=2 and σ=0.5. The lognormal distribution is not symmetric like the normal distribution, instead, it is skewed with a longer tail on the right side. To find the probability of an individual loss exceeding 10, we first convert the value 10 into its corresponding z-score using the transformation for a lognormal distribution.

To calculate the z-score, we use the formula:
Z = (ln(X) - μ) / σ

Where ln(X) is the natural logarithm of 10, μ is the mean (2), and σ is the standard deviation (0.5) of the logarithmic values of X.
Plugging in the values, we get:
Z = (ln(10) - 2) / 0.5

After calculating Z, we look up this z-score in the standard normal distribution table (or use a calculator) to find the probability that Z is greater than this value. This gives us the probability that an individual loss exceeds 10, which requires applying the properties of the lognormal distribution.