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#1
05-10-2011, 07:48 AM
 spencerhs5 Member Join Date: Jan 2007 Posts: 1,048
Negative Volatility

So if your stochastic equation is ds/S=.5375dt - .25 dz notice the minus sign for dz and you are asked for the bound of a confidence interval (soa #64 is an example) say .9 symmetric then your tw z's are +- 1.645. But it turns out the upper boundary is found using -1.645 because multiplying the two negatives together is what causes the addition in the exponent. Am I missing something?
#2
05-10-2011, 07:58 AM
 Scars Member CAS Join Date: Nov 2009 Studying for CAS Exam 5 Posts: 3,883

Quote:
 Originally Posted by spencerhs5 So if your stochastic equation is ds/S=.5375dt - .25 dz notice the minus sign for dz and you are asked for the bound of a confidence interval (soa #64 is an example) say .9 symmetric then your tw z's are +- 1.645. But it turns out the upper boundary is found using -1.645 because multiplying the two negatives together is what causes the addition in the exponent. Am I missing something?
For B-S Formula (d1 and d2), B-S Equation, confidence intervals, probabilities, and monte carlo estimates, you will use the absolute value of sigma.

For sharpe ratios, you will use sigma.

The exponent for the upper bound is:
$(.5375-\frac{|-.25|^2}{2})t + |-.25| sqrt{t} (1.645)$

If it said the stock was non-divided paying, and r = .10, the sharpe ratio would be:
$\frac{.5375-.10}{-.25}=.-1.75$

Make sense?

Last edited by Scars; 05-10-2011 at 08:05 AM..
#3
05-10-2011, 08:01 AM
 Scars Member CAS Join Date: Nov 2009 Studying for CAS Exam 5 Posts: 3,883

Quote:
 Originally Posted by Joe F Perhaps it's time for a more thorough explanation of why it's convenient to accept the idea of a negative value of sigma when calculating the Sharpe ratio. Suppose we encounter a contingent claim with the following process: $\frac{dV[S(t),t]}{V[S(t),t]}=\alpha_V [S(t),t]dt + \sigma_V [S(t),t]dZ(t)$ And suppose that during the exam we need to find V's Sharpe ratio. The easiest way to find the Sharpe ratio is: $\hspace{20} Sharpe Ratio = \frac{\alpha_V [S(t),t] - r}{\sigma_V [S(t),t]}$ But there is an alternative approach that works too. I'm not a proponent of this alternative approach, but here's how it goes: First determine whether $\sigma_V [S(t),t]$ is positive or negative. 1. If it is positive, then we can use the usual formula: $\hspace{20}\sigma_V [S(t),t] >0 \hspace{15} \rightarrow \hspace{15} Sharpe Ratio = \frac{\alpha_V [S(t),t] - r}{\sigma_V [S(t),t]}$ 2. If it is negative, then we make a couple of adjustments to the formula: (i) we use the absolute value of $\sigma_V [S(t),t]$ in the formula, and (ii) we multiply through by -1. This gives us: $\hspace{20} \sigma_V [S(t),t] < 0 \hspace{15} \rightarrow \hspace{15} Sharpe Ratio =- \frac{\alpha_V [S(t),t] - r}{|\sigma_V [S(t),t]|}= \frac{\alpha_V [S(t),t] - r}{\sigma_V [S(t),t]}$ Notice that #1 and #2 above produce the same result. That's why, to me, it doesn't seem worth the time to determine the sign of $\sigma_V [S(t),t]$ and then make adjustments (i) and (ii). You can get the same Sharpe ratio by simply using the formula with the given value of $\sigma_V [S(t),t]$.
This may help.
#4
05-10-2011, 08:02 AM
 Scars Member CAS Join Date: Nov 2009 Studying for CAS Exam 5 Posts: 3,883

Quote:
 Originally Posted by Joe F The usual form for a bond, P, uses a negative sign in front of the volatility term: $\frac{dP}{P}=\alpha dt - q dZ(t)$ The negative sign above means that when we define the Sharpe ratio as follows, the Sharpe ratio will be positive: $SharpeRatio = \frac {\alpha - r}{q}$ Now we are just using g in place of P: $\frac{dg}{g}=[r(t)+0.09] dt - \beta dZ(t) \hspace{15} \rightarrow \hspace{15} SharpeRatio = \frac {[r(t)+0.09] - r}{\beta}$
Because it seems natural to wonder what this would imply for bonds (in regards to Sharpe ratios).
#5
05-10-2011, 09:01 AM
 dickmwong Member SOA Join Date: Mar 2011 College: University of Hong Kong; London School of Economics Posts: 212

volatility is just the standard deviation, so it must be a positive number.
The negative sign is considered in the Sharpe ratio only. If there is another asset with a positive coefficient for dZ(t), that means these two assets are perfectly yet negatively correlated.
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