Binomial Distribution (binomial model inspired by Merton)
Hi I want to do the following and I hope that anyone of you can help me out.
In short, I have a portfolio consisting of 10 companies, whereas their asset values are driven by a common economic background factor u and the assets in question are correlated. Now I want to calculate the probability that e.g. 3 of 10 companies default simultaneously.
In order to do so, I choose a mixed binomial Merton model. whereas the probability that a given number of companies default is given by:
whereas u is the common economic background factor and a standard normal random variable.
C(i) is given by:
whereas I substituted r with μ
Then I entered the following definitions:
V1 = 100; D1 = 70; μ1 = 0.2; σ1 = 0.1; V2 = 110; D2 = 75; μ2 = 0.25; σ2 = 0.2; V3 = 95; D3 = 72; μ3 = 0.1; σ3 = 0.2; V4 = 120; D4 = 100; μ4 = 0.4; σ4 = 0.3; V5 = 120; D5 = 110; μ5 = 0.2; σ5 = 0.2; V6 = 100; D6 = 65; μ6 = 0.25; σ6 = 0.12; V7 = 99; D7 = 85; μ7 = 0.3; σ7 = 0.05; V8 = 125; D8 = 76; μ8 = 0.2; σ8 = 0.3; V9 = 105; D9 = 100; μ9 = 0.23; σ9 = 0.02; V10 = 100; D10 = 75; μ10 = 0.2 ; σ10 = 0.2; T = 5; ρ = 0.5; \[ScriptU] = 0.25; Evaluating Subscript[C, i] C1 = (Log[V1/D1] + (μ1 - 0.5*σ1^2)*T)/(σ1*Sqrt[T]); C2 = (Log[V2/D2] + (μ2 - 0.5*σ2^2)*T)/(σ2*Sqrt[T]); C3 = (Log[V3/D3] + (μ3 - 0.5*σ3^2)*T)/(σ3*Sqrt[T]); C4 = (Log[V4/D4] + (μ4 - 0.5*σ4^2)*T)/(σ4*Sqrt[T]); C5 = (Log[V5/D5] + (μ5 - 0.5*σ5^2)*T)/(σ5*Sqrt[T]); C6 = (Log[V6/D6] + (μ6 - 0.5*σ6^2)*T)/(σ6*Sqrt[T]); C7 = (Log[V7/D7] + (μ7 - 0.5*σ7^2)*T)/(σ7*Sqrt[T]); C8 = (Log[V8/D8] + (μ8 - 0.5*σ8^2)*T)/(σ8*Sqrt[T]); C9 = (Log[V9/D9] + (μ9 - 0.5*σ9^2)*T)/(σ9*Sqrt[T]); C10 = (Log[V10/D10] + (μ10 - 0.5*σ10^2)*T)/(σ10*Sqrt[T]);
Now i am not sure How to calculate the individual probabilities and add them up.
Would you suggest first to integrate the terms on the right hand side of the integral or is it possible to enter the whole function in Mathematica considering both the Normaldistribution and the Binomialdistribution?
I would be very thankful for any advise.