Investments, tenth edition

Figure 5.4 

The normal distribution with mean 10% and standard deviation 20%.

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 Early descriptions of the normal distribution in the 18th century were based on the outcomes of a “binomial 

tree” like that of the newspaper stand. This representation is used in practice to price many option contracts, as we 

will see in Chapter 21. For a demonstration of how the binomial distribution quickly approximates the normal, go 

to   www.jcu.edu/math/isep/Quincunx/Quincunx.html  .  

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  C H A P T E R  

5

  Risk, Return, and the Historical Record 

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multilayered relationship. But when securities are normally distributed, the statistical rela-

tionships between returns can be summarized with a straightforward correlation coeffi-

cient. Thus we need to estimate only one parameter to summarize the dependence of any 

two securities. 

 How closely must actual return distributions fit the normal curve to justify its use in 

investment management? Clearly, the normal curve cannot be a perfect description of real-

ity. For example, actual returns cannot be less than  2 100%, which the normal distribution 

would not rule out. But this does not mean that the normal curve cannot still be useful. 

A similar issue arises in many other contexts. For example, birth weight is typically evalu-

ated in comparison to a normal curve of newborn weights, although no baby is born with a 

negative weight. The normal distribution still is useful here because the SD of the weight 

is small relative to its mean, and the likelihood of a negative weight would be too trivial 

to matter.  

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   In a similar spirit, we must identify criteria to determine the adequacy of the 

normality assumption for rates of return.

   

 

 

  

 Suppose the monthly rate of return on the S&P 500 is approximately normally distrib-

uted with a mean of 1% and standard deviation of 6%. What is the probability that 

the return on the index in any month will be negative? We can use Excel’s built-in func-

tions to quickly answer this question. The probability of observing an outcome less than 

some cutoff according to the normal distribution function is given as NORMDIST(cutoff, 

mean, standard deviation, TRUE). In this case, we want to know the probability of an 

outcome below zero, when the mean is 1% and the standard deviation is 6%, so we 

compute NORMDIST(0, 1, 6, TRUE)  5  .4338. We could also use Excel’s built-in  standard  

normal function, NORMSDIST, which uses a mean of 0 and a standard deviation of 1, 

and ask for the probability of an outcome 1/6 of a standard deviation below the mean: 

NORMSDIST( 2 1/6)  5  .4338.