Corrections Q1 The expected value computed should be divided by P(1.2X>2000) to give conditional expected value as insurer would usually be notified about only those claims exceeding the deductible amount, hence the need to compute conditional expectation automatically. Although one may possibly get full marks for giving unconditional expected value as well. Q5 i) It should be 0.96*0.03 + 0.03*0.73
For question 6 iii, can we say that the answer is [1 - P(X > 1.64, Y > 2.33) ] - P(X < −1.64, Y < −2.33)? The first term in the square brackets would give P(X
Corrections
Q1
The expected value computed should be divided by P(1.2X>2000) to give conditional expected value as insurer would usually be notified about only those claims exceeding the deductible amount, hence the need to compute conditional expectation automatically. Although one may possibly get full marks for giving unconditional expected value as well.
Q5 i) It should be 0.96*0.03 + 0.03*0.73
Any chance you'd be able to go through the CB2 2024 Multiple Choice/some of the qs?
Sir when will you upload cb2 solutions?? Looking forward to that
We wont be releasing for CB2 this session.
Sir please make videos on cb1 mocks
Please upload cb2 solutions as well
For question 6 iii, can we say that the answer is [1 - P(X > 1.64, Y > 2.33) ] - P(X < −1.64, Y < −2.33)? The first term in the square brackets would give P(X
Thanks sir
Hello sir, what would be the expected cutoff?
expected passing marks??
For Q7, going through Cov(Xt,et-k) should have been easier.
Yes, that approach could have also been used & would have been easier to comprehend for many.
Paper B solutions??
We wont be releasing that.
Paper b?
We wont be releasing that.