J = 54, with probability 0.75 = 0 with probability 0.25 * 0.5 = 50 * U with probability 0.25 * 0.5, where U is uniform over (0,1) No idea re the mix distribution. For instance, I guess you have N as counting variable for loss/no loss \[ P[N=0]=0.75\\ P[N=1]=0.25 \] then \( P[J=54]=P[N=0]=.75\) since saying you have 54 it is as saying you did not experience any loss. Let s now be \(L\) r.v. for the size of the Loss. Then we should have \[ P[J=0]=P[L=50|N=1] \cdot P[N=1]=\frac{1}{100}\cdot 0.25 \] since \(L \approx U(0,100)\) isn't it? The last row eludes me...
Yes, that's right. Not quite. Since the Junior Loan repays zero whenever the loss is 50 or more, we want: \(P[J=0]=P[L>=50|N=1] \cdot P[N=1]=\) 0.5 x 0.25 For losses of less than 50 (which happens 0.25 x 0.5 of the time), the Junior Loan repays between 0 and 50. This can be represented by 50U where U is Uniform over (0,1).
Hi I'm just stuck with the last part of Q9 (i). In the examiner's report it says S = 50 * U with prob 0.25 * 0.5, shouldn't it be S = 50 - 50*U (i.e. in the event that a loss occurs, and the loss is greater than 50, the amount returned to senior loan holders are 50 reduced by the excess of the loss over 50)? Please help.
U means a U(0,1) RV. So 50 - 50U = 50(1-U) = 50U, since the amount returned is uniformly distributed between 0 and 50.
