Non-Systematic Variance

y2alk

New Member
Subscriber
Hi,

Can someone explain me understand the below step

If the portfolio were equally weighted, then the nonsystematic variance would
be

upload_2016-1-29_11-13-30.png
 

jairamjana

Member
Hi,
I will explain.. First this deals with Sharpes Single Index Model which is similar to typical regression line equation with R(m) = Returns from market, being an independent variable, dependent variable R(p) = Returns from portfolio.

R(p) = alpha(p) + Beta(p) * R(m) + error term(p)

We can find the E[R(p) | R(m)] based on the above but your question deals only with the Variance part. So we convert the equation accordingly.

Var(R(p)) = Var(B(p) * R(m)) + Var(e(p))
(Remember Variance of constant variable is zero . Therefore Variance of alpha is zero.)

Var(R(p)) = B(p)^2 * Var(R(m)) + Var(e(p))
Total Risk = Systematic Risk + Unsystematic Risk/Diversification Risk


B(p)^2*Var(R(m)) is always prevalent irrespective of size because the terms are from Beta of portfolio and the Market Variance.

Var(e(p)) is really firm specific and also Var(e(p)) being independent has a expected mean of zero, the Variance of this term decreases as number of assets increases - law of averages (i.e. Diversification reduces this component)

Now n being total assets in portfolio p. The above assertion can be proved if we have a equally weighted portfolio and therefore each asset will be (1/n) of total assets and i = 1,2,3.......n

Therefore
summation operator [Var(1/n*e(i))] = summation operator [(1/n)^2 * Var(e(i))]
Summation operator in this case means "Sum the values of Var(e(i)), starting at e(1) and ending with e(n)."
We can then say 1/n * 1/n = (1/n)^2 and we can bring 1/n outside the summation term and make the formula more elegant.
= (1/n) * {summation operator [(1/n) * Var(e(i))]}

Note that [ summation operator (1/n) * Var(e(i)] is nothing but average of sum of variance of error term i.e (var(e(i)) with a dash above: see below) ]
eqn.png

Notice as n increases to infinity the variance of error terms becomes negligible.. In Calculus terms this is a horizontal asymptote and this is how a graph looks..
systematic_risk.gif

Hope its clear.
 
Last edited:

y2alk

New Member
Subscriber
Thanks for the detailed explanation Jairam.

Can you tell me how the average of sum of variance of error term becomes 1/Var(e(i))

Is there any generalization used here? that was my question.
 

jairamjana

Member
Its not
1/Var(e(i))
Its
summation (1/n*Var(e(i)) )
And yes this is like

summation (1/n*(X(i)) ) = Average of Random Variable X(i)
 

jairamjana

Member
Ok I understood your question ...
upload_2016-1-29_11-13-30-png.401


This is not Variance to the power (-2) ... That minus like symbol is a dash actually.. That's how a mean of variable is symbolized..

This is the correct way as I explained in first post
eqn.png
 

jairamjana

Member
Haha.:D. your question sounded general I thought you were meaning explain the whole equation... Very well at least if someone wants to understand the derivation that elaborate post will be useful...
 
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