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For the common benefit

PostPosted: Tue Aug 05, 2008 11:56 am
by Muble
Interview with major IB. Here is a few I can remeber in random order:
1) Q 5.16 in mj
2)Q5.5 in mj. Answer in mj was not satisfactory as they wanted sth more optimal in time .....sorry, personally i did like your solution ;)
3) Solve f_x (x) = -( x* f(x) ) ^2
4) integrate sin^2(x)
5) integrate various other stuff all coming down to IBP
6) if S_t is GBM for which value of a is S_t^a, a martingale
7) Variation of Q3.23 in mj
8) Show that BM has nowhere differentiable paths ( !!!!!!!!!!) ( he couldnt show it either )
9) Why is the fact that BM has finite mean square variation finite, important ?
10) You play bridge ( 13 cards in 4 palyers ) and I accidently turn over a card and it is the ace of spades. What is the probaiblity that i have another ace? Suppose now that you only see it is an ace ( not of spades but anything). Is the probability of having two aces higher or lower than before?
11) Show put call parity
12 ) The equity market exhibits a vol skew and the interest rate a smile . Why ? Why not both a smile?
13) Have you ever worked with multi thread programming ? Functional languages ? NO ? *sigh*
14)Q3.2 in mj
Cant rember any more now but there were many....

PostPosted: Wed Aug 06, 2008 12:35 am
by mj

re 5.5, clearly you needed to do the follow-up question!

Interesting to note that you got two from our top ten (11 and 14)

did you pass the interview?

PostPosted: Wed Aug 06, 2008 8:15 am
by Muble
I should have Mark, but thankfully i managed to work sth it out in the interview.
I didnt pass and I should have because in retrospect, it wasnt that difficult ( given your book too ).
What do you think about 8) and 13) ?

PostPosted: Wed Aug 06, 2008 8:44 am
by Muble
oh! I also got #10 out of the top 10 i.e. Q7.10 right after the *sigh* in 13)
and here is one more i just rembebered ( and messed up)

What is the largest number that can not be written as a linear combination of 6,9, and 20 ?

PostPosted: Wed Aug 06, 2008 11:33 am
by crnt2
I'd be interested to know a good answer for 12.

For your question above, I get the answer to be 43 by a combination of logic and trial and error. My reasoning:

If you can find six consecutive numbers that you *can* write as a sum of 6, 9 and 20 then you can get any number above the smallest in that sequence by adding multiples of six. So the problem becomes:

"Where does the first sequence of six numbers that *can* be written as such a sum occur?"

5 obviously can't, so we proceed by checking the numbers of the form 5n+6:

11 can't
17 can't
23 can't
29 can

So there is a possible sequence 24, 25, 26, 27, 28, 29 which *can* be written as such a sum. But actually 25 *cannot* be written as such a sum (easy to see) so we start checking the numbers of the form 6n+25

31 can't (we can't use 20 because 11 fails, and 31 isn't a multiple of 3)
37 can't (same reasoning as above)
43 can't (same reasoning as above)
49 can

So there is a possible sequence 44, 45, 46, 47, 48, 49

44 = 20 + 4 x 6
45 = 5 x 9
46 = 2 x 20 + 6
47 = 20 + 3 x 9
48 = 8 x 6
49 = 2 x 20 + 9

So the answer is 43.

Is there a quicker way? That took me about 10 minutes to do without any pressure, so I wouldn't like to get it in an interview!

PostPosted: Wed Aug 06, 2008 11:37 am
by mj

44 = 4*6+20
45 = 5*9
46 = 2*20+6
47 = 3*9 + 20
48 = 8*6
49 = 2*20 + 9

any bigger number can be found by adding 6 to these

43 is not divisible by 3 so you must use 20

23 is not divisible by 3 so must use 2*20

3 is clearly not on.

So 43 is non constructible and we are done.

How to find the answer: any multiple of 3 bigger than 3 can clearly be done. 20 is congruent to 2 mod 3. So we would expect most numbers to be doable with 0 1 or 2 multiples of 20.

PostPosted: Wed Aug 06, 2008 11:42 am
by mj
oops cross post

PostPosted: Wed Aug 06, 2008 11:45 am
by mj
12) is a bit weird -- last time i looked they both exhibited skew and smile!

PostPosted: Wed Aug 06, 2008 4:21 pm
by Muble
For the "6,9,20" the way I (with hints ) solved it was pretty much
what mj and crnt2 suggest: First look at 6,9 and observe that all multiples of 3 except for 3 itself may be represented. Then by adding the 20 you observe that 60 is an upper bound for the number we are looking for being a multiple of 3 and of 20. And then write down the numbers 4 to 60 and start crossing out .( and in fact observe that you can stop at 44 for the reason you both mention.)

I was told that there are faster ways to do this......

For the vol question the argument was sth like :

people buy options as insurance. These people usually have long position in stocks ( as short positioning was recently introduced ). Hence they are interested in moves of the stock out of the money and hence the skew.

On the other hand in FX or IR markets long or short positions are as common. Hence pushing demand up both in and out of the money and hence a smile (????)

Mark , Im still waiting for tap in the back for 8) I think it was an unfair question
and having a math background, my ego was hurt. :roll:

PostPosted: Wed Aug 06, 2008 4:30 pm
by crnt2
Disclaimer: I have basically zero knowledge of the foundations of stochastic calculus or how you get to any of the results, so don't expect rigour from me!

You could argue heuristically along the lines:

W(t+dt) - W(t) is distributed normally with mean 0 and variance dt

Therefore W(t+dt) - W(t) = sqrt(dt) * N(0,1)

Therefore (1/dt) * (W(t+dt) - W(t)) = (1/sqrt(dt)) * N(0,1)

Which clearly doesn't converge to anything nice as dt -> 0.

Do you think that would have sufficed, or were they looking for an actual proof?

PostPosted: Wed Aug 06, 2008 4:42 pm
by Muble
I like your argument and I have no idea what they expected!

I tried to do the actual proof which kind of argues the the left Dini derivative is +\infty and the right is -\infty a.s. a.e but could not.. in fact as I checked later, the set where this happens is not even measurable. What you show is that it contains a set of probability 1

PostPosted: Thu Aug 07, 2008 5:31 am
by mj
i can't give you a short sweet proof. A good reference is first 20 pages
of Rogers and Williams.

PostPosted: Tue Aug 19, 2008 3:58 pm
by pussinboots
Regarding how to prove Brownian motion is not differentiable, crnt2's proof I think is short and sweet, as 1/sqrt(dt)) * N(0,1) converges to infinity in distribution.

Alternatively, it follows immediately from the fact that Brownian motion has non-zero quadratic variation, since this implies infinite variation, which in turn implies nowhere differentiability.

PostPosted: Tue Aug 19, 2008 8:31 pm
by mj
crnt's proof is good for showing that it's not a differentiable at a given point but that's much simpler then nowhere differentiable.

infinite first variation shows that it's not differentiable everywhere but that's not the same as nowhere differentiable. eg a function could be differentiable at one point only and have infinite first variation.

PostPosted: Tue Aug 19, 2008 9:09 pm
by pussinboots
Yeah, that's true. That didn't occur to me. But then again, having just looked up a proof for it, I doubt that is something you'd be expected to reproduce in an interview situation.