| Another definition of Risk |
There are a jillion definitions of "Risk" and I don't hardly like any of them.
>Don't tell me you're talking about risk again!
I know! I know! I've done it several times before. (Search here for "risk".)
However, I think many would like to think the following:
If stock A is "riskier" than stock B, then you'd expect:
- The probability of getting large returns is higher for stock A
- The probability of getting large losses is also higher for stock A
You'd like to be rewarded for taking a larger "risk", knowing full well that there's also a greater possibility of getting large losses.
>Yeah, so why call it "another definition of risk"? Why not stick with existing definitions like ...?
Like downside risk or standard deviation or value at risk or ... ?
>And they are ...?
Did I mention that you can search ... here?
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Anyway, I was thinking about such a definition and decided it shouldn't be called "risk" (which should imply only losses), but should be based upon both criteria noted above.
That is, the probability of larger gains as well as larger losses.
For example, which of the two stock return distributions would you think is ... uh, "riskier"? >With what definition of "risk"?
>In that case, I'd say the red stock is "riskier" ... but you really should give it another name.
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Which is riskier?
>You mean g-riskier, right? In that case, I'd say red ... that's Microsoft. But I coulda told you that without any g-risk mumbo-jumbo.
Well, yes but ...
>Besides, you're talking history, not future. I suggest you take a look at this.
Thanks ... but don't you think you'd like to avoid investments that are risky unless there's a reasonable chance of making BIG bucks?
>Investments that are risky? Define risky.
Okay, I mean investments that have a higher probability of losing money.
>Higher than what?
Higher than, say, some risk-free rate (like maybe 4%) or maybe higher than some Index, like the S&P500, or maybe ...
>Yeah, I'll go with that. I'd be prepared to accept a greater risk of loss, compared to the S&P, provided there was a good chance of getting a greater return.
Okay, then you'll like g-Risk. It compares the probability of getting a larger return (than some benchmark) to the probability of getting a smaller return.
For example, Value at Risk or VAR considers the probability of having a return less than some benchmark return.
We want to consider BOTH: less than something and greater than something so maybe we should consider a Ratio.
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>Don't you have to assume some kind of distribution of returns?
Yeah. In VAR it's normal to assume a ... uh, normal distribution. If we continue with that "normal" assumption, but do the Ratio thing, we'd get something like 
... as we did, once upon a time, in a Risk / Reward tutorial. Notice that the integral in the denominator is the probability of getting less than some return. It's subtracted from "1" to get the probability of getting "greater" than something. >And if the returns aren't normally distributed?
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Indeed, we're interested in probabilities of getting less than something (since both integrals in numerator and denominator are animals of this ilk).
That poses the following problem:
| How to generate a mathematical distribution that mimics an actual return distribution vis-a-vis the probabilities of getting less than something |
In other words, we need a mathematically defined distribution that mimics quantiles for some actual, historical distribution.
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>Quantiles?
Yes. Consider a random variable R. We want to have a 70% probability that R is less than something. What's that something? If the distribution satisfies Pr[R < ??] = 0.70
>Huh?
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I'm thinking.
| Another Distribution |
We look carefully at historical returns of some asset and extract the Mean m and Standard Deviation s.
Suppose we begin with the Normal distribution as a first approximation to the historical return distribution.
We'll call the Normal distribution N(r) where r is our random variable
and assign to N the Mean m and Standard Deviation s.
Our modified distribution is a perturbation of N. We'll call the deviation k.
Our "modified" distribution is then D(r) = N(r) + k(r).
To be a distribution, we'll need: ∫ D(r) dr = 1 ... the integration being over the range of r, namely -∞ to +∞.
That gives:
[0] ∫ k(r) dr = 0
... since ∫ N(r) dr = 1.
Then we'll want m
= ∫ r D(r) dr
= Mean[ D(r) ] = Mean[N(r) + k(r)]
= ∫ r (N(r) + k(r)) dr
= m + ∫ r k(r) dr
... the integration from -∞ to +∞.
That gives us the "first moment" of our perturbation:
[1] Mean[k(r)] = ∫ r k(r) dr = 0
Now on to the "second moment":
We'll want
∫ r 2 D(r) dr
= ∫ r 2 (N(r) + k(r)) dr
= m2 + s2 + ∫ r 2 k(r) dr
... since the 2nd moment for the Normal distribution is Mean2 + Standard Deviation2.
Let's call the historical moments that we want to mimic: μ1, μ2, μ3 ...
The above equation then reads: μ2
= m2 + s2 + ∫ r 2 k(r) dr
Then we have:
[2] ∫ r 2 k(r) dr
= μ2 - m2 - s2
Now on to the "third moment" ...
>Wait! What are we doing anyway?
We're trying to match the various moments of our modified distribution to the actual moments of the historical distribution.
Remember: If we extract the Mean and Standard Deviation from historical returns, that would completely define a Normal distribution.
We couldn't require of this Normal distribution that it match higher order moments.
That's why we're introducing a "modified" distribution. That extra term, k(r), will be defined so as to incorporate those additional moments.
>Those additional moments being the moments of the historical stuff, right?
Right. Let's continue.
Then we have:
[3] μ3 = ∫ r 3 D(r) dr
= ∫ r 3 N(r) dr
+ ∫ r 3 k(r) dr
= m3 + 3ms2 + ∫ r 3 k(r) dr
... since the 3rd moment for the Normal distribution is m3 + 3ms2.
>That 3rd moment for a Normal distribution ... I guess I take your word for it, eh?
Okay, consider this: for our Normal distribution where N(r) = e-(1/2) { (r - m)/s }2 / sqrt(2π)s
∫ rn N(r) dr = 1 / sqrt(2π)s ∫ rn e- { (r - m)/s }2/2 dr = 1 / sqrt(2π) ∫ (m + st)n e- t 2/2 dt ... putting r = m + st.
Expanding: (m + st)n = Σ nCz m n-z sztz ... where z goes from 0 to n and nCz are the binomial coefficients.
Then:
[4] μn
= ∫ r n D(r) dr
= Σ nCz m n-z sz
{ 1 / sqrt(2π) ∫ t z e- t 2/2(r) dr }
+ ∫ r n k(r) dr
... the integration being from -∞ to +∞.
We now have:
| [A]
∫ r n k(r) dr
= μn
- Σ nCz m n-z sz
Pz
|
>zzzZZZ
Don't you see?
We find the moments of the actual distribution, from historical data, and we know the values of Pz,
so we can determine the moments of k(r) .
That way we can modify the assumption that the actual return distribution is Normal. We just add our k(r) . 
>Huh? We know Pz?
Sure. Just look 'em up in a book. In fact, we already know that:
P0 = 1 ... true of all distributions
P1 = 0 ... the Mean = 0 for a "standard" normal distribution
P2 = 1 ... since the Standard Deviation = 1
P3 = 0 ... since all odd powers give a zero integral
>So what are the moments of this k-modification?
The first few are:
∫ r 0 k(r) dr = 0 ... that's [0], above
∫ r 1 k(r) dr = 0 ...see [1], above
∫ r 2 k(r) dr = μ2 - m2 - s2 ... see [2]
∫ r 3 k(r) dr = μ3 - m3 - 3ms2 ... see [3]
>What happened to μ0 and μ1?
N, the Normal component of our modified distribution, D, will take care of them.
It's them higher moments that we want to mimic by introducing our k-modification and ...
>But how do we calculate μn, the moments of the historical returns?
A good question.
>Okay, suppose we know those moments. Now how do we construct k from its moments?
Another good question.
Click to continue ...