Fixed Income interview

bump*

does anyone have brainteasers/puzzles to share?

i interviewed for a derivatives job..swaps

they asked the diff between Zero curve and yield curve...
you should know how to explain bootstrapping...
also know a little about FRA.

asked me about delta of options...if my option is at the money...how much of the underlying do i need to have to hedge..do i go long short? assuming im initailly long

what is my worst trade?

thats it..
be very political..i interview with 6 diff ppl in one day..you must remember that you need to adapt to EACH ONE!! they want to see if you fit...think about it constantly. You will meet the hotshot, the nerd, the bitch..adapt

thanks a lot. i appreciate the feedback

know a bit about the market...for brainteasers, well here's one.

I give you two envelopes, the one has twice the amount of money as the other. You pick one, and I give you the option to switch. do you switch or not?

That's not really a brainteaser as a math question though, right?

50% chance of 1/2x = 1/4x + 50% chance of 2x = x, so you get an expected value of 1.25x.

(x, of course, is the amount in your envelope)

so you'd switch?

and what if you'd picked the other envelope first?, and if you would switch, if i give you the opportunity to switch again what do you do?

and btw, most FI brainteasers are math based.

I believe I read this in Crack's book. You keep switching based on merkhet's EV calculation.

"You keep switching based on merkhet's EV calculation."

so you'd just sit there and continuously switch back and forth? does that make any sense at all?

If I'd picked the other envelope first, the calculation is the same since the envelope I picked in the first situation would then be unknown and would have the same percentages and payoffs tacked to it.

Since I would switch, if you gave me the opportunity to switch again... well, I'd know what was in both envelopes, so that'd depend on which envelope has more money in it. (Or perhaps I'm not understanding the second iteration of your question.)

Why would I keep switching back and forth?

Ahh... the exchange paradox. Reconsider the nature of your prior beliefs and the e(v) is actually zero. if you want to be smart, also mention that e(v) = e(utility) only when you are risk neutral.

Maybe I am being dumb, but it sounds like the expected value is being overthought. I wouldn't think it would matter whether you switch or not since there are only 2 envelopes, so wouldn't the second event (the switch) be exactly the same as the first (the pick)? Or am I missing something with the EV?

Oh, I see. I looked up exchange paradox from Sherminator's post... I had understood that I got to look inside my envelopes.

Interesting paradox.

Also, isn't this essentially the same question as that one brainteaser with three doors in it? If you choose one door, does that affect the probabilities associated with the other two doors? (I can't remember the name of the other brainteaser.)

"Why would I keep switching back and forth?"

you're saying you'd switch in the first instance. but your reasoning applies again post switch. so you'd keep switching always expecting 5x/4 in the other envelope if you've never looked at the initial value.

Revsly gets the answer correct, and if you are interested in trying to come up with probability distributions for unbounded sets, you can do some more research yourself.

"Also, isn't this essentially the same question as that one brainteaser with three doors in it? If you choose one door, does that affect the probabilities associated with the other two doors? (I can't remember the name of the other brainteaser.)"

the monty hall problem? no, this is very different.

the doors is like this, from wikipedia

"Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?"

What do you think?

"Oh, I see. I looked up exchange paradox from Sherminator's post... I had understood that I got to look inside my envelopes. "

Fine, you get the first envelope, and you look at its value. You are now given the option to switch, should you take it? and more relevant, what would you pay to switch?

Jimbo

As long as you have no idea which amount X and 2X... its the same situation, no? So even if you look and see $500, you wouldn't know whether that is the smaller (other has $1000) or the larger (other has $250). So its theoretically the same situation as not looking.

sure...so you look at see $500. The other has $250 or $1000 right? Do you switch then?

a brainteaser I've been asked numerous time "what the sum of numbers 0-100? and How did you get your answer"

that one is much too easy....100x50+50.

yes, but I've been asked it on more that 1 occasion so I thought I'd put it out there

fair enough. what others have you gotten?

I had a series of shot gun multiplication question such as 8x8x8 followed by 9x9x9...

-How much would you pay to roll a 6 sided die...I forgot the details it was an expected value question I got the correct answer though...

Good point. It shouldn't matter whether you look in the envelope or not since originally, prior to picking either envelope, you should have a EV of 1.5X.

As for the Monty Hall problem, you do always switch. Since the guy always reveals a goat for you, either (1) you've picked the car and switching leaves you with a goat, (2) you've picked a goat and switching gives you a car, or (3) you've picked a goat and switching gives you a car. 66 2/3% chance.

The sum thing is rather easy because you just group 100s until you are left with an ungrouped 50.

I'm still somewhat bothered by the exchange paradox. I think the disparity comes from perspective. Comparing both envelopes simultaneously versus comparing one envelope using the other envelope as a baseline changes something. Though I've no idea what that could posibly change.

I just thought of something. If you've heard these problems before, what's the point of the brainteaser? Doesn't it go from being an aptitude test to being a memory test?

"I'm still somewhat bothered by the exchange paradox. I think the disparity comes from perspective. Comparing both envelopes simultaneously versus comparing one envelope using the other envelope as a baseline changes something. Though I've no idea what that could posibly change."

you cannot make any claim about the nature of the distribution of the money in the envelopes.

I would switch because the EV of the new choice is higher (0.5*(250+1000)) than the 500.

I think something is wrong with that thinking... but I'm pretty sure the original Expected Value is exactly the same, so it doesn't matter I believe. E(X)=.5(X+2X)

Ok I think I figured it out. If you knew beforehand that the Letter contained 500, the original EV would become EV= .5(500) + .25(1000) + .25(250)= 625. So it is the same expected value as if you were to switch. Does that make any sense?

Did you happen to be interviewing for Citi or JPM?

I got the exact same questions (at both places)... although I was asked from 6^3 up to 9^3. Regarding the die; the question states that you get $1 for each unit value in the role... how much would you pay to play this game (anything less than $3.50).

"Ok I think I figured it out. If you knew beforehand that the Letter contained 500, the original EV would become EV= .5(500) + .25(1000) + .25(250)= 625. So it is the same expected value as if you were to switch. Does that make any sense?"

so you are saying the expected value of either envelope is 625, and if you are holding the $500, would you switch?

I think i need to get some envelopes and start doing this....

People are obviously over-complicating this problem. It makes no difference whether you switch. Don't look at it terms of what envelope you choose, look at it in terms of two envelopes sitting on the table. There is a 100% chance that one envelope has $X and a 100% chance that the other envelope has $2X. No matter which you choose, you have a 50% chance of getting $X and a 50% chance of getting $2X. Switching does not affect these probabilities.

If you want to put a dollar value on the problem, you can say that one envelope has $500. No matter what, the other envelope already has $250 or $1000 -- but they both can't be true. In this case, if you switch, you either have a 100% chance of losing $250 or a 100% chance of gaining $500 - not 50/50. But, you don't know which outcome will occur, so it makes no difference whether you switch.

Agree with b that switching doeesn't matter.

You know that your envelope contains either x or 2x, such that your expected return from switching is 1.25x (50%*1/2x + 50%*2x = 1.25x); however, regardless of whether you've been shown the amount of money in the envelope, you will not know the value of x (since $500 could represent x or 2x) and as such will have no incentive to switch.

I was trying to say you don't switch... or at least you are "switch neutral," meaning you don't care. I was using that EV calculation to show that using the original expected value it wouldn't matter whether you open the envelope or not, your Expected Values will always equal, meaning you are indifferent. There would be no reason to think switching would give you any advantage. Perhaps the way I said it before wasn't so clear; so to be clearer, I wouldn't be inclined to switch in either case. Yeah, perhaps breaking out some envelopes would have been helpful haha.

B -- I did interview for Citi but these questions were not from Citi....yup 3.50 was the correct answer....

Jimbo what do you trade????

I got asked bunch of options technical, literally got bombarded. Math teaser? Sqrt of 4000. I said near 64, actually 63.25.

I know this isn't the answer to the question, but...

.5(500) + .25(1000) + .25(250) != 625

might want to double check that.

Hahaha, you are right, I completely went downhill after deciding you shouldn't switch and I think my brain was desperately searching for justification and decided that was an answer. I need to sleep more and learn how to do division... I swear, its always the stupid errors that get me :)

girlytrader - interesting. I actually got the cubed questions at JPM (in the same shotgun format) and the die question at Citi.
I'm going to have to assume that you were interviewed by "Marty" of JPM then. I don't imagine many other people who would shotgun cubed questions (unless this is a new Street trend). As there was only 1 female at my superday... I just want to verify (b/c this would be a crazy coincidence it it were true) - please tell me that you don't go to McGill University.

brick - how did you solve that? Is there a trick?

I would just think round numbers and then narrow down. So you know 4000 is between 60^2=3600 & 70^2=4900, then it is closer to 60 than 70... you could keep narrowing (like it needs to be less than 65 since 65^2=4225, probably closer to 65 than to 60, so guess a little over halfway, like 63 or 64) or just take a guess at that point. I'm sure there is a better way, but this seems to work for me

well you could do mental square roots. but i think brick just did a quick guess based on intuition.

http://www.homeschoolmath.net/teaching/sqr-algorithm-why-works.php

Nearly same approach as Revsly. Figured 4000=40*100. Take out the 100 which is 10^2. 40 is between 6^2 and 7^2 and should be more on left side of 6.5 than right. And then just guess and multiply with with 10.

Ummm.... SQRT 4000 = 20 * SQRT(10) = 20* ~3.2 = ~ 64

Why are people using 65^2 = 4225?!

sherm thats a very good way to solve it fast.

I did not interview at JPM .... I said I interviewed at Citi but that was for a SA position and I was not asked any brainteasers

I was asked all these questions at the BB that I was an SA at and will be a FT trader at next July...sorry

Jimbo,

the envelope question: if you peek at it, doesn't it become subjective whether it is high or not? and then that usually affects a person's willingness to switch (or not).

jimbo, are you in FI?
can you explain to me the convexity adjustment for a CMS swap?
(seriously, thanks for your help)

i still don't get the envelope question.

i agree that in jimpos original question with only two envelopes switching is not profitable.

however, when you have three envelopes and have one with $500 in your hand I would switch if I knew that in the other two envelopes are $250 or $1000. I mean you either loose $250 or win $500, chances for both are 50%....aren't they?