问一道题：NO.PZ2020033002000010 [ FRM II ]

NO.PZ2020033002000010 问题如下 Assume ththe one-yeprobabilities of fault for the AAanBBB-ratebon are 1% an4%, respectively, anthe joint probability of fault of the two bon is 0.07%. Whis the fault correlation between the two bon? 0.03% 1.54% 78.91% The fault correlation cannot calculatewith the information provi B is correct.考点fault Correlation解析From Equation: p(A anB)=Corr(A,B)p(A)[1−p(A)]×p(B)[1−p(B)]+p(A)p(B)p{(\text{A anB})}=Corr{(\text{A,B})}\sqrt{p{(\text{A})}{\lbrack1-p{(\text{A})}\rbrack}}\times\sqrt{p{(\text{B})}{\lbrack1-p{(\text{B})}\rbrack}}+p{(\text{A})}p{(\text{B})}p(A anB)=Corr(A,B)p(A)[1−p(A)]​×p(B)[1−p(B)]​+p(A)p(, the fault correlation is Corr(A,B)=p(A anB)−p(A)p(B)p(A)[1−p(A)]×p(B)[1−p(B)]Corr{(\text{A,B})}=\frac{p{(\text{A anB})}-p{(\text{A})}p{(\text{B})}}{\sqrt{p{(\text{A})}{\lbrack1-p{(\text{A})}\rbrack}}\times\sqrt{p{(\text{B})}{\lbrack1-p{(\text{B})}\rbrack}}}Corr(A,B)=p(A)[1−p(A)]​×p(B)[1−p(B)]​p(A anB)−p(A)p(B)​=0.0007−0.01×0.040.01×(1−0.01)0.04×(1−0.04)=0.0154=\frac{0.0007-0.01\times0.04}{\sqrt{0.01\times(1-0.01)}\sqrt{0.04\times(1-0.04)}}=0.0154=0.01×(1−0.01)​0.04×(1−0.04)​0.0007−0.01×0.04​=0.0154. 题目是否还需要说明 AAanBBB-ratebon的违约概率均满足Bernoulli分布？