April 2010 Q9

[kalky]

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...

 

[Mike Lewry]

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.

Yes, that's right.

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?

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

The last row eludes me...

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).

 

[kalky]

tks a mil!

 

[Sendo]

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.

 

[Mark Mitchell]

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.