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Malz single factor model

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hi David,
Could you please explain the below mentioned sentence

Given m a realization of Ei less than or equal to Ki - Bim triggers default. As we let m vary from high to low values a smaller idisyncratic shock will suffice to trigger default

David Harper CFA FRM

David Harper CFA FRM
Staff member
Hi @saurabhpal49 Malz single-factor model is intermediate/advanced, I just want to "warn" you so that you do not expect to immediately grok it, as it's proven to be challenging . I copied below the latest version of my rendering of his example Malz 8.4. To grok the specific sentence you cite, IMO, is difficult without understanding the broader model (if you already get it, great!). Below I copied Malz example, which is specifically illustrated by my XLS further below.

To specifically answer your question, the meaning of "as we let m vary from high to low values a smaller idisyncratic shock will suffice to trigger default," is illustrated below: If you look at the market factor, m row, as it decreases from m = 0 to m = -2.33, notice the idiosyncratic (random) shock required to default (final row) is decreasing.

In this single factor model, the asset return is given by a(i) = β(i)*m + sqrt(1-β^2)*ϵ(i), where (m) is the single factor that represents the macro-economy and epsilon is the random standard normal shock. By specifying m = 1, the model simulations a downturn and, consequently a conditional distribution with a higher probability of default. I'll stop here because i don't want to overwhelm. I hope that's helpful, thanks!

Example 8.4 (Default Probability and Default Threshold) Suppose a firm has β(i) = 0.4 and k(i) = −2.33, it is a middling credit, but cyclical (relatively high βi). Its unconditional probability of default is Φ(−2.33) = 0.01. If we enter a modest economic downturn, with m = −1.0, the conditional asset return distribution is N(−0.4, sqrt[1 − 0.402]) or N(−0.4, 0.9165), and
the conditional default probability is found by computing the probability that this distribution takes on the value −2.33. That probability is 1.78 percent.

If we were in a stable economy with m = 0, we would need a shock of −2.33 standard deviations for the firm to die. But with the firm’s return already 0.4 in the hole because of an economy-wide recession, it takes only a 1.93 standard deviation additional shock to kill it.

Now suppose we have a more severe economic downturn, with m =−2.33. The firm’s conditional asset return distribution is N(−0.932, 0.9165) and the conditional default probability is 6.4 percent. A 0.93 standard deviation shock (ϵ(i) ≤ −0.93) will now trigger default [dharper: I disagree with this value]
This XLS is also here https://www.dropbox.com/s/7dnceu42dtpvzw7/P2.T6.Malz_Ch8_v2.xlsx?dl=0
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