I also got:

6 Months = .3982%
12 Months = 1.0461%
Maximum Prob when T = 20

However, It's probably a good idea to reword the last problem.  As it reads, it seems like the problem wants you to take into account if it goes insolvent inbetween time periods.  IE: The company would be insolvent if it went <0 during time 15 but then went >0 at time 20.  However, the way we are calculating it as a normal distribution - this would be counted as solvent. 

This could also be applied to part D, however it does say at the END of those time periods.  My suggestion is to remark somewhere that the company can only be determined insolvent at the END of the time period.

[NEAS: Yes, your re-wording is correct.]

I am also using different websites which gave me 1.047 percent.I am also in the list who agrees with you Wangxy.




Tim Manning New Zealand

Yes, Wangxy, I roughly agree with you, I am currently without excel, so i didnt get the optimal distribution, i had to use some website, which gave me 1.048 percent, but that falls in range with your 1.046%, so i would say that yes, i agree

 

and I also agree with Wolffollower, that to max t=20

Anyone getting 0.3982% after 6 month and 1.046% after 1 year for part D?

I get t=20.  I maximized the function -(200+10t)/40sqrt(t), which is the Z-value from the standard normal distribution representing the probability of being less than 0.

WF

I'm also getting an answer of t=2.  I've double checked for cancelling out an extra 10 and it seems correct to me.  Has anyone else come up with this answer? 

Thanks!

Ray is right on in helping everyone and I mean no disrespect, but I don't think part E is a MLE problem, it's just simple optimization.  Just in case anyone was confused by that.

   -    Dave

Ben, there is a hint for this in part B.  First of all, your variance at time zero is zero, as we are given the initial value.  The volatility of each month is independent, so in variance, this is additive.  For month one, there is only one month of volatility, so we square 40mil and take the square root.  For the second month, we would square 40mil for the first month, square 40mil for the second month, add the two months together and then take the square root of the sum.  Dot dot dot.  For the nth month, we square 40mil n times, add the n terms and take the root of the sum.  This leaves a cute, little formula

Variance(t) = 40mil² x t.  Thus: Standard Deviation(t) = 40mil sqrt(t)

RDH

please help to correct me if i'm wrong
we have Y-zero=200m, theta-zero=10m and var(et)=40^2
and the capital after one month is
Y-hat(1) = Y-zero + theta-zer0 which is equals to 210
so mean of capital after one month E(Y-hat(1)) = 210?
but what is the variance?

peat, I would suggest a cup of coffee and some reading glasses.

A random walk is a white-noise process (and thus a normal distribution) with a drift (a mean which increases linearly with time) and volatility (a much cooler discription for the standard deviation).  We are given the intial mean, its "slope" and the standard deviation with respect to time: 200m, 10m, and 40m.  It is straightforward to derive their equations from there to answer the first few questions.  Remember we are given the standard deviation, not the variance, so be careful when attempting to find the new standard devation for one and six months out.  The last two questions are fun, use the standard normal tables or excel to help find your answers.  I think the final question relates to maximum likelihood estimation, but we talk about that in the previous posts.

Other than that, I'd also suggest the beach and perhaps a picnic for the weekend, enjoy!

RDH