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Topic Title: Interview Questions
Created On Wed Aug 27, 03 05:30 PM

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GM
Member

Posts: 106
Joined: Aug 2002

Wed Aug 27, 03 05:30 PM

Here are some of the technical questions I was asked at a recent quant interview:

1. What is the limit of (x^2 + x)^0.5 - x as x tends to infinity?
2. Current stock price is 100. In the next period, it can go up to 150 with probability 0.9, or down to 90 with probability 0.1. How much is a one period call with strike 130 worth if the risk free rate is zero?
3. y'' + y' + y = 0
i) what is the general solution? [Answer: Ae^(alpha*t)]
ii) solve for alpha
iii) factorise out the real component of the solution
iv) do you recognise the resulting function?

Please show working so that I can check my logic!

Posts: 32
Joined: Apr 2003

Wed Aug 27, 03 08:03 PM

why do they demand ph. d's and then ask q's that would barely strain an undergrad?

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robertral
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Posts: 655
Joined: Mar 2003

Wed Aug 27, 03 08:05 PM

Quote
Originally posted by: GM

2. Current stock price is 100. In the next period, it can go up to 150 with probability 0.9, or down to 90 with probability 0.1. How much is a one period call with strike 130 worth if the risk free rate is zero?
q]

Answer to 2, i think, is C = exp(0)*[0.9*20 + 0.1*0] = 18

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Alainniala
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Wed Aug 27, 03 08:19 PM

Why are you using the probabilities given as being the risk neutral ones?

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robertral
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Posts: 655
Joined: Mar 2003

Wed Aug 27, 03 08:47 PM

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Originally posted by: Alainniala
Why are you using the probabilities given as being the risk neutral ones?

How else would you do it from the info given?

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Alainniala
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Posts: 23
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Wed Aug 27, 03 09:16 PM

up move: 1.5
down move: 0.9

--> Risk neutral probability = (1-0.9)/(1.5-0.9) = 0.17 = risk neutral prob of going up

and then use 0.17 and (1-0.17) as your probabilities...

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robertral
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Posts: 655
Joined: Mar 2003

Wed Aug 27, 03 09:23 PM

Cool, thanx, yeah just had a look at the Hull book on it.
I always forget the easier stuff when I'm looking at the harder stuff :S -> that's my excuse lol

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Posts: 27
Joined: Aug 2003

Thu Aug 28, 03 09:15 AM

if you find these difficult you are wasting your time applying for quant jobs.

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SauceDrinker
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Posts: 75
Joined: Nov 2002

Thu Aug 28, 03 11:01 AM

1. +0.5 multiply by conjugates.
2. 20/6.
3. See undergrad DE book.

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silverside
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Posts: 249
Joined: Jan 2003

Thu Aug 28, 03 11:38 AM

For 1) I would have done it by simplifying and getting the Taylor expansion of root(1+1/x) around 1; The constant term then falls out at 1/2.

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robertral
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Posts: 655
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Thu Aug 28, 03 12:38 PM

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Originally posted by: leibniz99
if you find these difficult you are wasting your time applying for quant jobs.

Who said I was applying for quant jobs? ;-)

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"WAAH EXACCLY I PACE!!!!!!!!!!!!!!!!!!!!!?"??..."WAAH EXACCLY I PACE!!!!!!!!!!!!!!!!!!!!!?"??

Posts: 27
Joined: Aug 2003

Thu Aug 28, 03 01:28 PM

Quote
Originally posted by: GM
Here are some of the technical questions I was asked at a recent quant interview:

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robertral
Senior Member

Posts: 655
Joined: Mar 2003

Thu Aug 28, 03 01:44 PM

I thought I'd have a go at answering them. It doesn't mean that I'm applying for quant jobs

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Edited: Thu Aug 28, 03 at 01:45 PM by robertral

Posts: 1887
Joined: Sep 2002

Thu Aug 28, 03 02:38 PM

on the ode problem, recall that the 'rates' in the exponential terms coincide with the roots of the polynomial over C... that's because if you put
f(t;r) = e^rt as your generic solution and then think of it as a function of r rather than t, this transformation turns differentiations into powers of r. So then you need the roots of

r^2 + r + 1

i.e. r = (-1 +/- i.sqrt(3)) / 2 and the solution has sine and cosine terms attached to an exponential decay etc.

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MinusTheBaboon
Junior Member

Posts: 9
Joined: May 2003

Thu Aug 28, 03 03:25 PM

1/
1/2 - multiply top and bottom (x^2 + x)^0.5 + x

So
((x^2 + x)^0.5 - x) ((x^2 + x)^0.5 + x)
--------------------------------------------------
((x^2 + x)^0.5 + x)

=

x^2 + x - x^2
------------------------
((x^2 + x)^0.5 + x)

=

x
-----------------------
((x^2 + x)^0.5 + x)

The denominator goes to 2x. x/2x->1/2

Posts: 1887
Joined: Sep 2002

Thu Aug 28, 03 03:40 PM

why work so hard when (1+x)^0.5 = 1 + x/2 + O(x^2) as x->0? (i.e. Taylor)

We have (x^2 + x)^0.5 = x . (1 + 1/x)^0.5 = x(1 + 1/2x + O(x^-2)) = x + 1/2 + O(1/x).

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Fools!
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Posts: 125
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Thu Aug 28, 03 03:54 PM

Quote
Originally posted by: robertral
Quote
Originally posted by: Alainniala
Why are you using the probabilities given as being the risk neutral ones?

How else would you do it from the info given?

Why wouldn't you? Risk neutral world valuation gives the same results as real world since the risk preference
is not involve. That is the method used for pricing derivatives using binomial trees anyway even for complex problems,
not for this straightforward one.

Also in q3 the CP has 2 complex roots so y=Aexp(k1*x) + Bexp(k2*x)