Negative Volatility

[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?

[Scars]

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:


If it said the stock was non-divided paying, and r = .10, the sharpe ratio would be:


Make sense?

[Scars]

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:



And suppose that during the exam we need to find V's Sharpe ratio. The easiest way to find the Sharpe ratio is:




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 is positive or negative.

1. If it is positive, then we can use the usual formula:


2. If it is negative, then we make a couple of adjustments to the formula: (i) we use the absolute value of in the formula, and (ii) we multiply through by -1. This gives us:


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 and then make adjustments (i) and (ii).

You can get the same Sharpe ratio by simply using the formula with the given value of .

This may help.

[Scars]

Originally Posted by Joe F

The usual form for a bond, P, uses a negative sign in front of the volatility term:



The negative sign above means that when we define the Sharpe ratio as follows, the Sharpe ratio will be positive:



Now we are just using g in place of P:

Because it seems natural to wonder what this would imply for bonds (in regards to Sharpe ratios).

[dickmwong]

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.