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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 = 100×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 α = 1.96
We find the extreme values from $100exp(μ ± ασ)
The lower limit is then V1 = $100 *  exp(0.10 − 1.96 * 0.2) = $100 * exp(−0.292) = $74.68. The upper limit is
V2 = $100 *  exp(0.10 + 1.96 × 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 ‘A’
 

 

Example

For the standard normal distribution, calculate the value of P(-1.87 ≤ Z ≤ 1.23) or P(|Z| ≤ 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 ≤ Z ≤ 1.23) or P(|Z| ≤ 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 ≤ Z ≤ 1.23) or P(|Z| ≤ 1.6) = P(Z ≥ 1.87) + P(Z ≥ 1.6) = 0.4693 + 0.4452 = 0.9145
Hence option ‘D’ is correct.

 





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