Can someone help with how to calculate the annualized probability of a loan default given:
70% probability of survival (30% default) over the next 20 months?
Edit:
I should have been more specific in my question. Actually, here is the problem. At month 10 into the loan, there is a probability of survival of 80%. At month 36, there is a probability of survival of 60%. I am trying to determine the annualized probability of default between these two months.
Thanks, John
$\endgroup$ 12 Answers
$\begingroup$It depends on what kind of default model you have. Are people more likely to default as they go into the loan, or is the probability the same regardless of where they are in the loan?
If the latter, then let's say there's some probability $x$ of default each month. Then the probability that it goes for $20$ months is $P_{surv}(20) = (1-x)^{20} = 0.7.$ Solving for $x$ gives $$x=1 - \sqrt[20]{0.7} \approx 0.017676.$$
Then, the probability of default for 12 months is
$$P_{def}(12) = 1-(1 - x)^{12} \approx 19.3 \%.$$
The same logic can be applied to a different default model. If $P_{surv}(10) = 0.8$ and $P_{surv}(36) = 0.6,$ then there's some probability of default $y$ in each of those months in between. So $P_{surv}(36) = 0.8(1-y)^{26} = 0.6,$ and proceed as above. Then
$$y = 1-\sqrt[26]{0.75} \approx 0.01100,$$
giving an annualized default rate of
$$P_{def}(12) = 1-(1 - y)^{12} \approx 12.5 \%.$$
$\endgroup$ 3 $\begingroup$Assuming a constant rate $x$ of default over each month, the rate of survival after $n$ months is $(1-x)^n$.
You are given $(1-x)^{20}=0.7$ and asked to compute $(1-x)^{12}=(0.7)^{0.6}$.
In the revised task, much depends on the model of the probability of default. If it is constant $x$ for all months from 11 to 36, then $0.8\cdot(1-x)^{36-10}=0.6$ and the result is again $1-(1-x)^{12}=1-(0.75)^{\frac6{13}}$, about 12.5%.
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