# Summation stability

Property of standard normal distributions

• The property states that, assuming there is a collection of normal distributions:
• The sum of the means of all the independent normal distributions form a normal distribution
• The sum of the variances of all the independent normal distributions form a normal distribution

Question: FRM Exam 2002
Consider a stock with an initial price of $100. Its price one year from now is given by S = 100Ã—exp(r), where the rate of return r is normally distributed with a mean of 0.1 and a standard deviation of 0.2. With 95% confidence, after rounding, S will be between 1.$67.57 and $147.99 2.$70.80 and $149.20 3.$74.68 and $163.56 4.$102.18 and $119.53 Solution C. Note that this is a two-tailed confidence band, so that Î± = 1.96 We find the extreme values from$100exp(Î¼ Â± Î±Ïƒ)
The lower limit is then V1 = $100 * exp(0.10 âˆ’ 1.96 * 0.2) =$100 * exp(âˆ’0.292) = $74.68. The upper limit is V2 =$100 *  exp(0.10 + 1.96 Ã— 0.2) = $100 * exp(0.492) =$163.56

Example

Let X be a uniformly distributed random variable between minus one and one so that the standard deviation of X is 0.577. What percentage of the distributions will be less than 1.96 standard deviations above the mean:
100%
97.5%
95%
Insufficient information provided

Solution

A
The answer requires understanding of distributions and standard deviation. The key is that every distribution has a standard deviation. However the number of standard deviations associated with different probabilities are different for each distribution. In this case 1.96 standard deviation represents a move of 1.12 or less. As the total distribution is defined as falling between minus one and one the correct answer
is â€˜Aâ€™

Example

For the standard normal distribution, calculate the value of P(-1.87 â‰¤ Z â‰¤ 1.23) or P(|Z| â‰¤ 1.6)?
0.5683
0.8794
0.7831
0.9145

Solution

D.
In the diagram given below, the area representing the region P(-1.87 â‰¤ Z â‰¤ 1.23) or P(|Z| â‰¤ 1.6) is shown below. The area will be from Z = -1.87 to Z = 1.6 and common area is from Z = 1.6 to Z = 1.23.
P(-1.87 â‰¤ Z â‰¤ 1.23) or P(|Z| â‰¤ 1.6) = P(Z â‰¥ 1.87) + P(Z â‰¥ 1.6) = 0.4693 + 0.4452 = 0.9145
Hence option â€˜Dâ€™ is correct.