Quantitative Interview questions and answers

[Guanting Liu]

The correct answers are discussed in later posts. I haven't updated my first posted for a while.

Check your calculation. It should be
(\ln a = \frac{1}{2} \ln (2 + a))
(a^2=2+a \Rightarrow a =2)
There are easier way to do this without using log but the idea is the same.

Still can't believe this was from 2007, when I was 12.
Graduated a math PhD and applying quant positions right now. Using this post as a question poll.
Just want to say thank you for establishing this post and keeping up the discussions.

 

[rangiku]

1st one

let x^x^x.. = 2
thus, x^(x^x^x...) =2
which can be expressed as x^2=2
therefore x= +/- sqrt(2)

 

[Lukee]

Still can't believe this was from 2007, when I was 12.
Graduated a math PhD and applying quant positions right now. Using this post as a question poll.
Just want to say thank you for establishing this post and keeping up the discussions.

Lol this is exactly how I feel reading posts on here from before I was born

 

[Andras]

After checking my calculation again, I have 1h 5m 27.27s. It's from 3927.272727... seconds. Probably you have the same number.

You have pencil and the answer sheet so people calculated mentally or scribbled anywhere they could . On average, you have 1.5-2 minutes for each question.

I am not sure if this is the correct solution, but my solution is built upon the fact that after 12 rotation of the hour hand, the minute hand has overtaken the hour hand 11 times. This means that 11*(time to overtake)=12 rotation of the hour hand. so the (time to overtake) = 12/11*(60 minutes) = 1h, 5.4545... minutes

 

[Quasar Chunawala]

Trick stochastic calculus question: what is the quadratic variation of [math]\int_{0}^{t}B(s)ds[/math] where [imath](B(t),t\geq 0)[/imath] is a standard brownian motion?

 

[ngnguyen]

Trick stochastic calculus question: what is the quadratic variation of [math]\int_{0}^{t}B(s)ds[/math] where [imath](B(t),t\geq 0)[/imath] is a standard brownian motion?

Modify the stochastic integral by making a change of variable as follows
[math]\int_0^tB_sds=\int_0^t\left(\int_0^sdB_u \right)ds=\iint_{0\le u\le s \le t}dsdBu =\int_0^t\left(\int_u^t ds \right)dB_u = \int_0^t(t-u)dBu[/math]
Then, apply the Ito isometry to find the quadratic variation
[math]\mathbb{E} \left(\left(\int_0^tB_sds\right)^2 \right)=\mathbb{E} \left(\left(\int_0^t(t-u)dBu\right)^2\right) = \int_0^t(t-u)^2du = \frac{t^3}{3}[/math]

 

[Quasar Chunawala]

Modify the stochastic integral by making a change of variable as follows
[math]\int_0^tB_sds=\int_0^t\left(\int_0^sdB_u \right)ds=\iint_{0\le u\le s \le t}dsdBu =\int_0^t\left(\int_u^t ds \right)dB_u = \int_0^t(t-u)dBu[/math]
Then, apply the Ito isometry to find the quadratic variation
[math]\mathbb{E} \left(\left(\int_0^tB_sds\right)^2 \right)=\mathbb{E} \left(\left(\int_0^t(t-u)dBu\right)^2\right) = \int_0^t(t-u)^2du = \frac{t^3}{3}[/math]

Since its an ordinary integral, I used the product rule [imath]\int u dv = uv - \int v du[/imath]. Then, you get:

[math] \int B_s \cdot 1 ds = sB_s|_{0}^{t} - \int_{0}^{t} u B_u = \int_{0}^{t} (t-u)dB_u [/math]

However, in this case the variance [imath]\neq[/imath] quadratic variation. The quadratic variation is zero. Let [imath] I_t = \int_{0}^{t} B_s ds [/imath]. By first principles:

[math]I_{t_{j+1}} = B_{t_0}(t_1 - t_0) + \ldots + B_{t_j}(t_{j+1} - t_j)[/math]

[math]I_{t_{j}} = B_{t_0}(t_1 - t_0) + \ldots + B_{t_{j-1}}(t_{j} - t_{j-1})[/math]

[math]
I_{t_{j+1}} - I_{t_{j}} = B_{t_j}(t_{j+1} - t_j)
[/math]

[math]
\sum_{j=0}^{n-1} (I_{t_{j+1}} - I_{t_{j}})^2 = \sum_{j=0}^{n-1} (B_{t_j})^2 (t_{j+1} - t_j)^2 \leq \max (t_{j+1} - t_j) \sum_{j=0}^{n-1} (B_{t_j})^2 (t_{j+1} - t_j)
[/math]

Taking expectations on both sides:
[math]
\mathbb{E}\left[\sum_{j=0}^{n-1} (I_{t_{j+1}} - I_{t_{j}})^2\right] \leq \max (t_{j+1} - t_j) \sum_{j=0}^{n-1} \mathbb{E}(B_{t_j})^2 (t_{j+1} - t_j)
[/math]

Since [imath]B_t[/imath] has almost surely continuous paths on any interval [imath][0,T][/imath], and continuous functions on a compact interval are bounded, so [imath](B_t)^2[/imath] is bounded on [imath][0,T][/imath]. There exists [imath]M[/imath] such that [imath]B_{t_j}^2 \leq M[/imath].

[math]
\mathbb{E}\left[\sum_{j=0}^{n-1} (I_{t_{j+1}} - I_{t_{j}})^2\right] \leq \max (t_{j+1} - t_j) M \sum_{j=0}^{n-1} (t_{j+1} - t_j) = \max (t_{j+1} - t_j) M T
[/math]

As the mesh-side approaches zero, the quadratic variation approaches [imath]0[/imath].

This also justifies heuristically [imath]dI_t \cdot dI_t = (B_t)^2 dt \cdot dt = 0[/imath].

 

[Paul Lopez]

Math rocks!

 

[Quasar Chunawala]

Math rocks!

I will suffix it - it rocks truly, if its an actual value-add to trading/structurers/trade management/any teams you are working with, and the solutioning makes intuitive sense and makes their life easy.

 

[techno_quant]

brain teaser?

Hi,

Not sure if anyone will ever ask this but my dad told me about this one and I thought I'd share:

You have an equation:

VIII - VII = VII

How do you change one of the "sticks" to make this correct?

V would be 2 sticks and I is 1 stick.

Lemme know what you think and if you can solve it.
:D

move the third stick of VIII to the minus to make it an equals sign? VII = VII = VII

 

[Lukee]

move the third stick of VIII to the minus to make it an equals sign? VII = VII = VII

How about VII - VII ≠ VII

 

[quantmenot]

1. Find \(x\) if \(x^{x^{x^{\ldots}}}=2\)

2. Find all real and complex root of \(x^6=64\)

3. The hour and minute hands of a clock meet at 12'oclock. When will be the first time they meet again ?

4. 3 points are randomly drawn on a circle. What is the probability of them being on the same semi-circle ?

5. A unit length is broken off into 3 pieces. What is the probability of them forming a triangle ?

6. Calculate \(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\ldots}}}} \)

7. There are 14 identical-looking balls. 13 of them have the same weight while one of them is heavier than the rest. What is the minimum times you can weight to identify the heaviest ball ? How do you generalize for n balls ?

Not to poop on the party, but I'd like to point out that question #1 is phrased poorly. As phrased, the solution would be (as @bob had suggested):

Let [math]y = x^{x^{x^{\ldots}}} = 2[/math] Then [math]2 = y = x^y = x^2 \iff x = \sqrt{2}[/math] (and what happens to the negative root? )

But, as it turns out, [math]\sqrt{2}[/math] also solves the following recurrence:

[math]x^{x^{x^{\ldots}}} = 4[/math] whence, as before, we arrive at [math]x^4 = 4 \iff x = \sqrt{2}[/math]

It follows that

[math]2 = \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}} = 4[/math]

In fact, it is true that [math]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}} = 2[/math]
but the proof is a little more subtle, and I'm going to leave it for someone on this thread to come up with (unless there's no answer in a day or two, in which case I won't be able to help myself..)

 

[ngnguyen]

Not to poop on the party, but I'd like to point out that question #1 is phrased poorly. As phrased, the solution would be (as @bob had suggested):

Let [math]y = x^{x^{x^{\ldots}}} = 2[/math] Then [math]2 = y = x^y = x^2 \iff x = \sqrt{2}[/math] (and what happens to the negative root? )

But, as it turns out, [math]\sqrt{2}[/math] also solves the following recurrence:

[math]x^{x^{x^{\ldots}}} = 4[/math] whence, as before, we arrive at [math]x^4 = 4 \iff x = \sqrt{2}[/math]

It follows that

[math]2 = \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}} = 4[/math]

In fact, it is true that [math]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}} = 2[/math]
but the proof is a little more subtle, and I'm going to leave it for someone on this thread to come up with (unless there's no answer in a day or two, in which case I won't be able to help myself..)

I agree that the question #1 is ill-posed: what [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}[/imath] means? And althought the answer seems corrrect, it is not rigorous.

Besides, I think [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}} \ne 4[/imath] because
[math]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...\color{red}{\sqrt{2}}}}} < \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...\color{red}{\sqrt{4}}}}} =... = \sqrt{2}^{\sqrt{2} ^{2}} = \sqrt{2}^{2} = 2[/math]

 

[quantmenot]

I agree that the question #1 is ill-posed: what [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}[/imath] means? And althought the answer seems corrrect, it is not rigorous.

Besides, I think [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}} \ne 4[/imath] because
[math]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...\color{red}{\sqrt{2}}}}} < \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...\color{red}{\sqrt{4}}}}} =... = \sqrt{2}^{\sqrt{2} ^{2}} = \sqrt{2}^{2} = 2[/math]

The expression [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}[/imath] is well defined (else, you should also take issue with the expression [imath]x^{x^{x^{...}}}[/imath]).

I'm not following what you mean by a "correct answer". My answer stands to illustrate the faults of the question to start out with.

My objection was that you cannot just claim that the recurrence [imath]x^{x^{x^{...}}} = 2[/imath] has a solution, any more than you could just claim that [imath]x^{x^{x^{...}}} = 4[/imath] has a solution, leading to the obvious contradiction that I described.

Additionally, I don't think the expressions [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots \color{red}{\sqrt{2}}}}}[/imath] and [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots \color{red}{\sqrt{4}}}}}[/imath] mean what you think that they do:

Firstly, it is certainly true that

[math]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}} \ne 4[/math]

Now, suppose that [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}} = 2[/imath] (as it does). Then, for example,

[math]
\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots \color{red}{\sqrt{4}}}}} = 2^{\color{red}{\sqrt{4}}} = 2^{\color{red}{2}} = 4 \ne 2
[/math]

The expression [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}}[/imath] is to be understood as follows:

Let [imath]x_0 = \sqrt{2}[/imath], and for [imath] n > 0[/imath] define [imath]x_n = \sqrt{2}^{x_{n-1}}[/imath]

Then

[math]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}} := \lim_{n \rightarrow \infty} x_n[/math]

One would have to prove, of course, that the limit exists, and is equal to [imath]2[/imath].

 

[Quasar Chunawala]

The expression [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}[/imath] is well defined (else, you should also take issue with the expression [imath]x^{x^{x^{...}}}[/imath]).

I'm not following what you mean by a "correct answer". My answer stands to illustrate the faults of the question to start out with.

My objection was that you cannot just claim that the recurrence [imath]x^{x^{x^{...}}} = 2[/imath] has a solution, any more than you could just claim that [imath]x^{x^{x^{...}}} = 4[/imath] has a solution, leading to the obvious contradiction that I described.

Additionally, I don't think the expressions [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots \color{red}{\sqrt{2}}}}}[/imath] and [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots \color{red}{\sqrt{4}}}}}[/imath] mean what you think that they do:

Firstly, it is certainly true that

[math]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}} \ne 4[/math]

Now, suppose that [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}} = 2[/imath] (as it does). Then, for example,

[math]
\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots \color{red}{\sqrt{4}}}}} = 2^{\color{red}{\sqrt{4}}} = 2^{\color{red}{2}} = 4 \ne 2
[/math]

The expression [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}}[/imath] is to be understood as follows:

Let [imath]x_0 = \sqrt{2}[/imath], and for [imath] n > 0[/imath] define [imath]x_n = \sqrt{2}^{x_{n-1}}[/imath]

Then

[math]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}} := \lim_{n \rightarrow \infty} x_n[/math]

One would have to prove, of course, that the limit exists, and is equal to [imath]2[/imath].

Hi, [imath] \sqrt{2}^{\sqrt{2}}> \sqrt{2}[/imath] and by the same logic [imath] \sqrt{2}^{\sqrt{2}^{\sqrt{2}}}> {\sqrt{2}^{\sqrt{2}}}[/imath], so it's a monotonically increasing sequence, and each term is smaller than 2. By monotone convergence theorem, the limit exists. And now, you can argue that since limits are unique, if the limit is [imath]y[/imath], [imath]\sqrt{2}^y = y[/imath]. So, [imath]2^y=y^2[/imath] which has only 1 root, [imath]2[/imath].

 

[quantmenot]

Hi, [imath] \sqrt{2}^{\sqrt{2}}> \sqrt{2}[/imath] and by the same logic [imath] \sqrt{2}^{\sqrt{2}^{\sqrt{2}}}> {\sqrt{2}^{\sqrt{2}}}[/imath], so it's a monotonically increasing sequence, and each term is smaller than 2. By monotone convergence theorem, the limit exists. And now, you can argue that since limits are unique, if the limit is [imath]y[/imath], [imath]\sqrt{2}^y = y[/imath]. So, [imath]2^y=y^2[/imath] which has only 1 root, [imath]2[/imath].

Almost! The first part is correct. Perhaps worth elaborating why each term is smaller than [imath]2[/imath], but you got the gist.

The second part isn't quite there. The problem is that the equation [imath]y = \sqrt{2}^y[/imath] does not determine the limit uniquely. In other words, it is true that the limit must satisfy [imath]y = \sqrt{2}^y[/imath], but there are other points that satisfy this equation as well (e.g., [imath]4 = \sqrt{2}^4[/imath], as before).

 

[Quasar Chunawala]

Almost! The first part is correct. Perhaps worth elaborating why each term is smaller than [imath]2[/imath], but you got the gist.

The second part isn't quite there. The problem is that the equation [imath]y = \sqrt{2}^y[/imath] does not determine the limit uniquely. In other words, it is true that the limit must satisfy [imath]y = \sqrt{2}^y[/imath], but there are other points that satisfy this equation (e.g., [imath]4 = \sqrt{2}^4[/imath], as before).

I see your point. Could I also argue that since [imath]\sqrt{2} \leq x_n \leq 2, \forall n[/imath], by the order limit theorem, the limiting value must fall in this interval.

 

[ngnguyen]

The expression [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}[/imath] is well defined (else, you should also take issue with the expression [imath]x^{x^{x^{...}}}[/imath]).

I'm not following what you mean by a "correct answer". My answer stands to illustrate the faults of the question to start out with.

My objection was that you cannot just claim that the recurrence [imath]x^{x^{x^{...}}} = 2[/imath] has a solution, any more than you could just claim that [imath]x^{x^{x^{...}}} = 4[/imath] has a solution, leading to the obvious contradiction that I described.

Additionally, I don't think the expressions [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots \color{red}{\sqrt{2}}}}}[/imath] and [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots \color{red}{\sqrt{4}}}}}[/imath] mean what you think that they do:

Firstly, it is certainly true that

[math]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}} \ne 4[/math]

Now, suppose that [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}} = 2[/imath] (as it does). Then, for example,

[math]
\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots \color{red}{\sqrt{4}}}}} = 2^{\color{red}{\sqrt{4}}} = 2^{\color{red}{2}} = 4 \ne 2
[/math]

The expression [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}}[/imath] is to be understood as follows:

Let [imath]x_0 = \sqrt{2}[/imath], and for [imath] n > 0[/imath] define [imath]x_n = \sqrt{2}^{x_{n-1}}[/imath]

Then

[math]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}} := \lim_{n \rightarrow \infty} x_n[/math]

One would have to prove, of course, that the limit exists, and is equal to [imath]2[/imath].

I want to say [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}[/imath] cannot equal to [imath]4[/imath].
Indeed, if we denote [imath]x_n = \sqrt{2}^{ \sqrt{2}^{...^{ \sqrt{2}}}}[/imath] ([imath]n[/imath] times), then
[math]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}} = \lim_{n\infty} x_n \tag{1}[/math]
For every [imath]n[/imath], we have [imath]x_n = \sqrt{2}^{ \sqrt{2}^{...^{ \color{red}{\sqrt{2}}}}}<\sqrt{2}^{ \sqrt{2}^{...^{ \color{red}{\sqrt{4}}}}}=..=2[/imath]
By consequence, [math]x_n \le 2 \tag{2}[/math] for all [imath]n \in \mathbb{N}[/imath]
From [imath](1),(2)[/imath], we deduce that
[math]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}} = \lim_{n\infty} x_n \le 2 \color{red}{< 4}[/math]

 

[quantmenot]

I see your point. Could I also argue that since [imath]\sqrt{2} \leq x_n \leq 2, \forall n[/imath], by the order limit theorem, the limiting value must fall in this interval.

Yes, you need to.

 

[quantmenot]

I want to say [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}[/imath] cannot equal to [imath]4[/imath].
Indeed, if we denote [imath]x_n = \sqrt{2}^{ \sqrt{2}^{...^{ \sqrt{2}}}}[/imath] ([imath]n[/imath] times), then
[math]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}} = \lim_{n\infty} x_n \tag{1}[/math]
For every [imath]n[/imath], we have [imath]x_n = \sqrt{2}^{ \sqrt{2}^{...^{ \color{red}{\sqrt{2}}}}}<\sqrt{2}^{ \sqrt{2}^{...^{ \color{red}{\sqrt{4}}}}}=..=2[/imath]
By consequence, [math]x_n \le 2 \tag{2}[/math] for all [imath]n \in \mathbb{N}[/imath]
From [imath](1),(2)[/imath], we deduce that
[math]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}} = \lim_{n\infty} x_n \le 2 \color{red}{< 4}[/math]

Not entirely following you here, but worth pointing out a few errors. For example, as I pointed out before,

[math]
\sqrt{2}^{ \sqrt{2}^{\ldots ^{\color{red}{\sqrt{4}}}}} = 2 ^ {\color{red}{\sqrt{4}}} = 2 ^ {\color{red}{2}} = 4 \ne 2
[/math]

Check out @Quasar Chunawala 's proof for the sequence being monotonously increasing, and bounded from below. For boundedness from above, observe that
[math]\sqrt{2} < 2 \implies \sqrt{2}^{\sqrt{2}} < \sqrt{2}^2 = 2 \implies \sqrt{2}^{\sqrt{2}^{\sqrt{2}}} < \sqrt{2}^2 = 2[/math]
etc. So that
[math]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}} := \lim_{n \rightarrow \infty} x_n \le 2[/math]