博迪第八版投资学第十章课后习题答案(10)

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13. The maximum residual variance is tied to the number of securities (n)

in the portfolio because, as we increase the number of securities, we

are more likely to encounter securities with larger residual variances.

The starting point is to determine the practical limit on the portfolio residual standard deviation, (e P), that still qualifies as a ‘well-

diversified portfolio.’ A reasonable approach is to compare 2(e P) to the market variance, or equivalently, to compare (e P) to the market

standard deviation. Suppose we do not allow (e P) to exceed p M, where p is a small decimal fraction, for example, 0.05; then, the smaller the value we choose for p, the more stringent our criterion for defining

how diversified a ‘well-diversified’ portfolio must be.

Now construct a portfolio of n securities with weights w1, w2,…,w n, so

that w i =1. The portfolio residual variance is: 2(e P ) = w122(e i)

To meet our practical definition of sufficiently diversified, we

require this residual variance to be less than (p M)2. A sure and

simple way to proceed is to assume the worst, that is, assume that the residual variance of each security is the highest possible value

allowed under the assumptions of the problem: 2(e i) = n2M

In that case: 2(e P ) = w i2n M2

Now apply the constraint: w i2 n M2 ≤ (p M)2

This requires that: n w i2≤ p2

Or, equivalently, that: w i2 ≤ p2/n

A relatively easy way to generate a set of well-diversified portfolios is

to use portfolio weights that follow a geometric progression, since the computations then become relatively straightforward. Choose w1 and a

common factor q for the geometric progression such that q < 1. Therefore, the weight on each stock is a fraction q of the weight on the previous stock in the series. Then the sum of n terms is:

w i= w1(1– q n)/(1– q) = 1

or: w1 = (1– q)/(1–q n)

The sum of the n squared weights is similarly obtained from w12 and a

common geometric progression factor of q2. Therefore:

w i2 = w12(1– q2n)/(1– q 2)

Substituting for w1 from above, we obtain:

w i2 = [(1– q)2/(1–q n)2] × [(1– q2n)/(1– q 2)]

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