接下來(lái)的任務(wù)是實(shí)際計(jì)算所謂 portfolio loss distribution 搬味。 為了給 credit derivatives 和CDO定價(jià)境氢，我們需要能夠計(jì)算這種投資組合損失分布portfolio loss distribution。 所以這是我們的第一個(gè)目標(biāo)碰纬。

• 如果我們對(duì)隨機(jī)變量M進(jìn)行條件處理condition on the random variable M萍聊，那么n個(gè)違約事件是獨(dú)立的。

• 特別是在這里給出了這些表達(dá)式悦析，給出的M條件下的違約概率寿桨。
  圖中公式來(lái)自這條：
  qi (t | M) = probability( Xi ≤ xi bar(t) | M)
  qi (t | M) = probability( [ai * M + [根號(hào)(1-a^2)] * Z_i] ≤ xi bar(t) | M)
  qi (t | M) = probability( Z_i ≤ { [xi bar(t) - ai * M] / [根號(hào)(1-a^2)] } | M)
  上面公式我們得出了圖中的公式，右手邊都是Constant强戴，這里的概率是 φ亭螟，累計(jì)概率CDF。
  

• 現(xiàn)在讓 P上標(biāo)n (l, t)骑歹，l 表示在時(shí)間t之前预烙， 總共有l(wèi)個(gè)投資組合違約的風(fēng)險(xiǎn)中性概率。

• 然后道媚，我們可以將t時(shí)間 的pl 寫成對(duì)于給定的M時(shí)扁掸，所有給定M情況下， p superscript n l of t 的負(fù)無(wú)窮到正無(wú)窮的積分最域。乘phi of MdM谴分。

We may then write pl of t as being the integral for minus infinity to infinity of p superscript n l of t given M times phi of MdM where phi is the standard normal PDF.

Now if you're wondering where this expression comes from, well, this is just a standard basic undergraduate expression for probability. In discrete form, you can imagine the following. Suppose we want to compute the probability that X is equal to little x. Well, a standard way of doing this is to consider another random variable y, and to sum over all possible values of y. So it's equal to the probability of X equals little x, and Y equals little y. And this is also equal to the sum over little Y of the probability that X equals little X. Given Y equals little Y by the probability that Y equals little Y. So this is a standard expression you probably all seen before in your undergraduate probability class. We're just using this here and. In density form rather than discrete probability mass function form. So M here takes the place of y. So we have our m here, taking the place of y over here, and so instead of a summation, we have an integral. And we're integrating with respect to m instead of summing over the y values here. So this is standard, so we can now write the probability of l default, and just to be consistent, I should have put a superscript n there. So the probability of l default out of the n names by time t, is given to us by this integral here. So the next task is going to be how do we compute this quantity? We know phi of M, it's just the standard normal density. So, we need to compute this quantity here. We can compute this quantity, then we can evaluate this integral numerically, and therefore compute this risk-neutral probability function here. So let's focus on how to compute this expression here. So, in fact, we can easily do it using an iterative procedure, and the iterative procedure will work as follows. So the first thing we're going to do let's initialize. We're going to have to run a couple of for loops here. So let's initialize our quantities first of all, we're going to set p 0. Given M to be equal to 1 minus q1 of M. We're going to set p1, given M to be equal to q1 of M. And we're going to set p2 given M to be equal to P3 given M all the way up to Pn given M. We're going to set all of that equal to 0. Now, what I'm doing here is, I'm dropping the dependence on n and t here. So, I don't want the slide to get too cluttered. If I did I'd have an n in all of these values here. And I'd have a t in here, and so on. But that's just going to get really cluttered. So, I won't do that. So, I'm dropping the dependence of these quantities on n, and t. So, now we're going to do the following. We're going to run the following for loop, we're going to say. For i equals 2 up as far as n. And, for j equals 1 to i. I am going to update these quantities here that i have already initialized. In particular I am going to say.
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P, j given M so this is the probability of j defaults, conditional on M. So this is going to represent the probably of j defaults in the first i names. So there, there are two ways we can get j defaults in the first i names. We could have j minus 1 defaults in those first i names, and having the ith name defaulting. Which is represented here or we could have had j defaults in those first i names. And then having the ith name not defaulting. So that's the end of this for loop. We must also handle the, the case of 0 defaults. So we must also update that. We get the probability of 0. Given M is equal to whatever the previous value was or the value in the previous iteration. Times 1 minus qi of M. In other words, what I'm really doing here. So this is the end of the, of the algorithm. What I'm really doing here is I'm lining up the names. I'm ordering them from say, 1 all the way up to to nth names. I know the risk neutral probability given n for all names. And I'm going through these names these credits in order starting with 1, all the way up to value capital N. I'm doing that via these two for loops here, and I'm updating the probabilities at each step. So, as I said here, so if I'm at this point, for some value of i and some value of j, what I'm doing is the following. I'm now going to update the probabillity of the j defaults. In the portfolio based on the first i names. So, the way j defaults can occur among the first i names is that j minus 1 defaults occur in the first i minus 1 names. Remember this value here is the value left over from the previous iteration. Which considers the names from 1 up to i minus 1, so I get this equals the probability of j minus 1 defaults in the first i minus 1 names. Times the probability that the ith name defaults, plus the probability of j defaults in the first i minus 1 names, by the probability that the ith name does not default. And so I can just run through these two four loops to compute pN l of t, given M. So in other words, at the very end once I get up to , N, i equals capital N and j equals i at the end of this procedure I have the probabilities that I want. And now at this point, we can perform a numerical integration on the right hand side of 2 to compute Pl of t. So this is Pl of t here. I now have computed all of this using that iterative procedure, and I can do numeric integration to calculate this quantity here. And there should be a super script in there as well. If we assume that the notional ai. And the recovery rate ri are constant across all credits, then the loss on any given credit will be either 0 or a times 1 minus r. Therefore knowing the distribution of the number of defaults is equivalent to knowing the distribution of the total loss and the reference portfolio. And that's because l losses on that case is equivalent to a loss of l times a 1 minus r and the portfolio of bonds and so knowing one implies the other. So this, the assumption of constat Ai's and Ri's actually simplifies the calculations, and we're going to be assuming that through out here. But I will mention that this assumption is easily relaxed. We could actually get by without it.