Scaling Margin Period of Risk for Linearly Declining Exposure

Summary

The document examines how to scale the margin period of risk when exposure declines linearly to zero. It compares a continuous exposure integral, which gives half the constant-exposure amount, with a discrete variance expression involving squared exposure levels. The author asks how that discrete approximation is derived and whether it comes from a Taylor expansion.

The post provides the formulas being compared and a proposed variance calculation, but it does not include an answer or establish that the proposed moments are correctly formulated. It is therefore useful as a focused question about continuous versus discrete approximations in exposure risk, rather than as a complete derivation. Readers should verify the definitions of the random variable, its integration limits, and whether the discrete expression represents variance or a sum of squared exposures before applying it to collateral or risk calculations.

Key ideas

• For exposure declining linearly to zero, the continuous integral over the margin period is half the corresponding constant exposure amount.
• The document contrasts that result with a finite discrete sum of squared exposure levels.
• The author asks for the derivation of a closed-form discrete approximation and whether it is a Taylor expansion.
• The proposed moment expressions are not validated or explained in the document.

Tags

# Margin period of risk and scaling (MPoR)


# Margin period of risk and scaling (MPoR)












I'm analyzing the formula to approximate the Margin Period of Risk (MPoR) for linearly linearly decreasing to zero exposure.

Given the MPoR at $\tau$ one can evaluate the continious total exposure at $$\int\limits_0^{\tau} \frac{\tau - u}{\tau} du = \frac{\tau}{2}.$$

So, the MPoR for linearly decreasing exposure is half the constant one (makes sense so far).

However, then there is discrete approximation as of Var as: $$\left[\left(1 - \frac1n \right)^2 + \left(1 - \frac2n \right)^2 + ... + \left(1 - \frac{n-1}{n} \right)^2 \right] = \frac n3 \left(1 - \frac1n \right)\left(1 - \frac{1}{2n} \right).$$

I've calculate the Var for $f(x) = 1 - \dfrac{u}{\tau}$:

- $\mathbb{E}(u) = \frac{u^2}{2} - \frac{u^3}{3\tau}$

- $\mathbb{E}(u^2) = \frac{u^3}{3} - \frac{u^4}{3\tau}$

- $\mathbb{Var}(u) = \frac{u^3}{3} - \frac{u^4}{3\tau} - \frac{u^4}{4} + \frac{u^5}{3\tau} - \frac{u^6}{6\tau ^2}$.

How can the approximation be obtained? it does not seem to be some simply Taylor series approximation.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.