Let's say you are about to put on a trade and it will leave an unhedgable risk position on your book for a month. And let's say you are very sure the unhedegable risk looks like a gaussian distribution with a width sigma (and no fat tails - very unlikely but let's do a thought experiment here).You should obviously charge for the risk, but do you charge an amount proportional to sigma or sigma squared?That is, if you had trades of size X or 2X, would you charge double the amount for the unhedgeable risk element in the 2X trade or four times the amount? Assume you are nowhere near risk limits, it is beginning of the trading year, etc.(There are arguments for both cases)

Given the distribution, its value to you is _entirely_ dependent on your risk preferences and utility function. Given any reasonable value, it should be possible to construct a utility function that will give exactly that value. If you are rational (in the classical economic sense) and risk-averse over dollar-value outcomes, there will be some restrictions on the feasible values, but not very strong ones.Similarly, how you scale with size is going to depend on your risk-preferences, too, as well as whether you are going to hit risk-limits/get margin calls.So, without thinking rather carefully about how you feel about the risk, as well as what the risk is likely to be, it is really impossible to answer your question.

Actually I think if you use a valid utility function, which would have to be negatively convex, then in the limit of a small risk position you will find the answer is that the certainty equivalent is linear in variance. This would suggest the answer to the question is more like "you charge four times as much if you double the position size".But I suspect traders would lean towards charging some number of standard deviations of the risk. This kind of behavior could not in general come from a utility function.

Suppose you are an expected utility maximizer (and lots more utilities are possible than this, even while being "reasonable") over dollar outcomes, then the certainty equivalent will be:E[u(X)] = \int u(x) \phi(x) dxAssuming that the PDF and the utility are analytic around the mean, a, and all the relevant integrals converge OK, you are going to getE[u(x)] = u(a) + 0.5 u''(a) \sigma^2 + ... + (1/n!) D^n u(a) m_n + ...where m_n is the nth central moment. Even if your distribution is Gaussian, the higher even moments are NOT zero - m_4=3\sigma^4 - though they are going to be a function of the variance. Convexity needs u''<0, but doesn't stop your higher derivatives having values which mean the certainty equivalent can depend on higher powers of \sigma. So, no, even for small bets, the certainty equivalent is not necessarily linear in variance even in the very special case where the payoff distribution is Gaussian for sure.

If the risk is sufficiently small, the gaussian would become increasingly like a delta function, so would pick out the lowest moments.For any reasonably behaved utility function you would clearly get expected value plus first order correction linear in variance, plus smaller contributions from higher order. And obviously this is in the limiting case.But let's get back to the main idea: Traders: would you charge 4 times the risk penalty for a trade twice as big? Or would it be more like twice the risk penalty?This is a trader forum, and I am looking for a practical answer, not a uniqueness/existence type answer.More than 50 traders have already answered this question so I guarantee it is possible.I think Paul Wilmott is right that there are too many mathematicians in finance!!

My point was not mathematical: you simply can't answer the question without thinking about your risk appetite. Not the scaling; not the absolute price; none of it.And to try to answer the question by asking others about their risk aversion is a curious thing to do. Their risk-aversion has no bearing on the question: yours is what matters (and your client/employer, if it is not all your own money).I think most traders would be likely to agree that one of the biggest issues with unhedgeable risk is your risk of being wrong about the distribution. That means that assuming your distribution is Gaussian is an extremely dangerous thing to do: it is the kind of mistake a mathematician might make, though, if he or she was not experienced in finance

Dude - the entire point of the question, which you have missed, is to ask others about their risk appetite. That is why more than 50 traders have answered the question, and zero of them have said "I can't answer it because I don't know my utility function!"And yes, the distribution is obviously key, which is why I said:"And let's say you are very sure the unhedgable risk looks like a gaussian distribution with a width sigma (and no fat tails - very unlikely but let's do a thought experiment here)."That is the *assumption*.I can see I am wasting my time here.

I'd look at it as an investment opportunity in terms of the Sharpe-ratios.The Sharpe-ratio would be approx unchanged if you charge twice as much for doubling your position and standard deviation.However even if the risk is normal, you can not indefinitely double the position and the charge because at some point your risk of loss will be of the same magnitude as the free capital and when that happens frictional costs will start to reduce your ‘real’ Sharpe-ratio and the relationship will no longer be linear i.e. you’ll only be willing to take marginal risk at increasing marginal premiums. It is possible a trader will still see linear relationship for large positions if he has an asymmetric payoff i.e. gets the upside but not the downside (apart from being fired).

QuoteOriginally posted by: erstwhileTraders: would you charge 4 times the risk penalty for a trade twice as big? Or would it be more like twice the risk penalty?Four arguments:1) Marginal Charge for Marginal Trade: Up to the risk limit and to a first-order analysis, the marginal cost or risk seems constant. Assume you did one trade like this of size X and charging a risk premium of c*X. What if the original counterparty wanted to double the size of the trade? Or, what if a second counterparty also wanted the same trade? Would you charge 3*c*X for the additional trade (i.e., a total of 4*c*X for a total of 2X position size)? And if a third trading opportunity arose, would you charge 5*c*X (for a total of 9*c*X for a total of 3X position size)? This seems unlikely.2) Opportunity Costs (Foregone profit): If one accepts this trade, then other trades will need to be foregone or scaled down to stay within the total risk limit. Depending on profitability and capacity of those other trades, the result is a linear or slightly higher than linear function of the charge of this hypothetical tarde. That is, as this trade consumes more and more of the trader's risk budget, the trader must foregone more and more of the best alternative opportunities. The exact increase in the charge is a function of the set of expected returns for other trades. This will be steeper than linear, but I doubt it will be O(X^2).3) Information Asymmetries: That the counterparty wants X or 2X (or nX) provides some information about the counterparty's expectations for returns. If the counterparty knows something that the trader doesn't, the counterparty's interest in greater size is an indicator of potentially greater expected returns to the counterparty and potentially worse losses to the trader. The more the counterparty wants to trade, the more wary the trader should be and the higher the charge.4) Market Price Sensitivity: A savy trader charges what the market will bear. If the counterparty wants to double the size, the trader might propose a much higher price to see if the counterparty will pay (with hand-waving rationale of increasing risk, opportunity costs, liquidity fears, risk-averse boss, blah blah blah to justify the higher price). Then a bit of bid/ask negotiation will find the right price.My personal choice is the reverse order of the above. I'd like to determine the market price sensitivity, then the information asymmetry risks, and then my opportunity costs before apply the linear marginal cost argument.Of course, all these arguments are meaningless if the counterparty is a valued client and the trader must (at all costs) avoid losing the client by overcharging for this one 2X trade.

I tend to agree with Traden4Alpha, in particular his last 2 points (from my experience). In these situations, from the perspective of a market-maker anyway, i turn the problem of what spread to charge with such exotic / unhedgable risk on its head. Let's assume you are getting paid for (as opposed to getting given) this risk. Rather than think about how far from "model fair value" you are happy to get paid, you should think about where you would be happy to buy this as a risk-reducing trade and spread/charge a bid-offer from there. Especially if you have already been paid for some of this risk prior to this, a trader who knows it is unhedgable should really be willing to have a better bid than that which he would have hypothetically shown the first time it traded, and if this next clip is in double size, the resulting offer for the next clip of risk is more likely to resemble 4x, but thats partly because the level at which you would buy the risk back is higher (so effectively skewing the bid-offer higher). Never underestimate the information you get by missing deals like this - it tells you where the market is comfortable pricing this risk, as opposed to where you individually might be content but are probably missing some information vital to your modelling distribution.