MAFS522 – Quantitative and Statistical Risk Analysis
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MAFS522 – Quantitative and Statistical Risk Analysis Topic Two – CreditRisk+ This is an industrial code that applies actuarial mathematics to calculate the loss distribution of a portfolio It requires a limited amount of input data and assumptions. It uses • as basic input the same data as required by the Basel II internal rating system. A loan is understood as a Bernuolli random variable (default or no default over a given time horizon). No profits or losses from rating migrations are considered. 1 It provides an analytic-based portfolio approach for rapid and un- • ambiguous calculations of the loss distribution (no simulations are needed). Efficient numerical technique, like the Fast Fourier trans- form, and analytic approximation method, like the saddle point ap- proximation method, can be applied to hasten the computation. – Unambiguity of the distribution is important (in particular, infor- mation on the tail of the distribution) since simulation based port- folio approaches usually fail to give agreeing answers (due to low probabilities of default). – Rapid calculations are helpful when comparative statistics (“what if” analyses) are performed: running different scenarios for the parameters. The resulting portfolio loss distribution can be expressed as a sum of independent compound negative binomial random variables. 2 Limitations Poisson approximation (multiple defaults may be possible) – assuming • the expected probabilities to be small. Though highly unlikely, possible paradoxes may be created, say, calculated losses > sum of exposures since multiple defaults of single obligor is allowed. Limited range of default correlation • – produces only positive and usually moderate levels of default cor- relation among 2 obligors sharing a common risk driver. Recovery rates and defaults are assumed to be independent. • Challenge Can the model bring out one of the most important credit risk drivers: concentration? 3 Fundamentals of CreditRisk+ – mixture Poisson distribution No modeling for default event: The reason for a default of an obligor • is not the subject of the model. Instead the default is described as a purely random event, characterized by a probability of default. Stochastic probability of default: The probability of default of an • obligor is not seen as a constant, but a randomly varying quantity, driven by one or more (systematic) risk factors. The distribution of the default intensity is usually assumed to be a gamma distribution. 4 Conditional independence: Given the risk factors, the default of oblig- • ors are independent. Only implicit correlations via risk drivers: Correlations between oblig- • ors are not explicit, but arise only implicitly due to common risk factors which drive the probability of defaults. A linear relationship between the systematic risk factors and the default probabilities is assumed. Using exposure bands: Group the exposures in the portfolio into • bands. Replace each exposure amount LA of obligor A by the nearest integer multiple of L, where L is the basic unit of exposure. The distribution of losses is derived from the probability generating func- tions, which can be used if the losses (count of defaulting obligors or dollar amount) take discrete values. 5 Probability generating functions GK(z) The probability generating function (pgf) of a discrete non-negative integer- valued random variable K is a function of the auxiliary variable z such that the probability that K = k is given by the coefficient of zk in the polyno- mial expansion of the probability generating function. K ∞ k GK(z)= E[z ]= P [K = k]z , kX=0 where z can be complex with z 1 (to guarantee convergence of the | | ≤ infinite series). The pgf of the Poisson random variable N with parameter α is n α(z 1) α ∞ (αz) G (z)= e − = e− , N n! nX=0 which observes the following probability mass function of N: e ααn P [N = n]= − . n! 6 For a Bernuolli random variable Y with probability p G (z) = (1 p)z0 + pz =1+ p(z 1). Y − − The pgf contains all the information to calculate the probability mass function of the associated non-negative integer valued random variable N, where 1 ( ) P [N = n]= G n (0). n! N Conditional pgf: G (z )= E[zN ]. N |· |· Given G (z x) where x has distribution F (x), then N | GN (z)= GN (z x) dF (x). Z | 7 Other properties Let K and K be a pair of independent non-negative integer-valued • 1 2 random variables. The pgf of the sum K1 + K2 is simply the product of the two pgf’s since E[zK1+K2] = E[zK1]E[zK2]. This stems from the property that for any given pair of independent random variables X1 and X2, E[f1(X1)f2(X2)] = E[f1(X1)]E[f2(X2)] for any functions f1 and f2. Here, we choose f1 and f2 to be the exponential function. G (z)= G (zn) for any natural number n. • nY Y For example, take n = 3, we have G (z) = P (3Y = 0)z0 + P (3Y = 3)z3 + P (3Y = 6)z6 + 3Y ··· = P (Y = 0)z0 + P (Y = 1)z3 + P (Y = 2)z6 + ··· = ∞ P (Y = k)z3k. kX=0 8 Variance formula for Y in terms of GY or ln GY 1 ∞ k 1 Y 1 2 Y G′ (z)= kP [Y = k]z = E[Yz ] and G′′ (z)= E[(Y Y )z ] Y z z Y z2 − kX=0 so that G (1) = E[Y ] and G (1) = E[Y 2] E[Y ]. Y′ Y′′ − Hence, σ2 = E[Y 2] E[Y ]2 = G (1) + G (1) G (1)2. Y − Y′′ Y′ − Y′ Note that GY (1) = E[1] = 1 and GY′ = GY (ln GY )′ and GY′′ = GY′ (ln GY )′ + GY (ln GY )′′ so that the derivatives of GY and ln GY are related by 2 GY′ (1) = (ln GY )′(1) and GY′′ = [(ln GY )′(1)] + (ln GY )′′(1). 2 An alternative representation of σY is 2 σY = (ln GY )′′(1) + (ln GY )′(1). 9 Input data for the model For an obligor A in a loan portfolio, we designate P to be the ex- • A pected probability of default of A (output of a rating process). e Write v as the potential loss for A upon default. The expected loss • A for A is given by e ǫA = PAvA. e To work with discretized losses, wee fix a loss unit L and choose a • positive integer vA as a rounded version of vA/L. To compensate for the error due to rounding, we adjust the expected default probability e ǫA vA PA = = PA. vAL vAL! e e We then work with PA and the integer-valued loss vA. 10 For example, suppose we take vA = $160, 000 and PA = 0.2%, then ǫ = $160e , 000 0.2%. e A × We take L = $100, 000 so that the rounded exposure in unit of L is 2. The adjusted expected default probability is $160, 000 0.2% P = × = 0.16%. A 2 $100, 000 × Pgf of number of defaults Since a default event is considered as a Bernuolli event, the pgf for a single obligor is F (z) = (1 P )+ P z. A − A A For the whole portfolio, we have F (z)= ∞ P (n defaults)zn. nX=0 11 Example Rounding of exposure to units of L Obligor A Exposure ($) Exposure Round-off Band j (potential loss (in $100,000) exposure given default) (in $100,000) vA vA/L vA 1 150,000 1.5 2 2 2 460,000e e 4.6 5 5 3 435,000 4.35∗ 5 5 4 370,000 3.7 4 4 5 190,000 1.9 2 2 6 480,000 4.8 5 5 Out of the 6 obligors, 2 to Band 2, 1 to Band 4, and 3 to Band 5. ∗ Rounding up to the nearest integer is adopted so that the round-off exposure as integer multiple of L always starts at 1 for the sake of convenience. 12 Allocation of 350 obligors into 6 bands vj number of obligors ǫj µj 1 30 1.5 1.5 2 40 8 4 3 50 6 2 4 70 25.2 6.3 5 100 35 7 6 60 14.4 2.4 vj = common exposure in band j in units of L ǫj = expected dollar loss in band j in units of L µj = expected number of defaults in band j We have ǫj = vjµj. Say, µ4 = 6.3 means the expected number of defaults among 70 obligors in band 4 is 6.3 over a given time horizon, and the potential dollar loss in band 4 is 4 6.3 = 25.2. × 13 Default events with fixed default rates As a consequence of independence between default events, the probability generating function for the whole portfolio is F (z)= F (z)= [1 + P (z 1)]. A A − YA YA Taking logarithm on both sides, and ln[1 + P (z 1)] P (z 1), we A − ≈ A − have ln F (z) P (z 1). ≈ A − XA As a Poisson approximation, we take µ n P (z 1) µ(z 1) ∞ e− µ n F (z) e A A − = e − = z , ≈ n! P nX=0 m where µ = E[1A] = PA = µj = expected number of default XA XA jX=1 events over a given time horizon from the whole portfolio. Here, 1A is the indicator default variable for obligor A. 14 If the probabilities of individual defaults are small, though not necessarily equal, the probability of realizing n default events in the portfolio over a given time horizon is e µµn P(n defaults) = − . n! This is the well known Poisson distribution with single parameter µ. The distribution has only one parameter, the expected number of • defaults µ.