Tower Research Capital Interview Questions 2026

Let \(T\) be the start time measured in minutes after 2:00 pm, so \(T\sim\mathrm{Uniform}[0,60]\) and the bell is at \(T=30\). For a uniform continuous distribution, probabilities are relative interval lengths. A singleton has length zero, hence

\[\Pr(T=30)=\frac{0}{60}=0.\]

However, \(30\) belongs to the sample space \([0,60]\). Probability zero is not the same as logical impossibility: the particular value is a legitimate outcome even though a continuous draw assigns it zero probability. By contrast, \(\{T<0\}\) is empty within the defined sample space and is impossible.

Within two minutes of the bell means \(28\le T\le32\), an interval of length four, so

\[\Pr(|T-30|\le2)=\frac{32-28}{60}=\frac1{15}.\]

Including or excluding the two endpoints does not affect this probability because each endpoint is itself a zero-probability singleton.

All eight inclusion patterns have prior probability 1/8. Only four can produce the announcement P:

Therefore \(P(\text{announce }P)=\frac{1}{8}+\frac{1}{16}+\frac{1}{16}+\frac{1}{24}=\frac{7}{24}\). For sound to be present and P to be announced, the hidden state must be \(\{P,S\}\) or \(\{P,S,I\}\), yielding joint probability \(\frac{1}{16}+\frac{1}{24}=\frac{5}{48}\).

Bayes' rule now gives

\(P(S\text{ present}\mid\text{announce }P)=\frac{5/48}{7/24}\)=\(\frac{5}{14}\).

The answer is not simply \(P(S\text{ present}\mid P\text{ present})=\frac{1}{2}\). The announcement is a random selection from what is present: in exhibitions with several features, each individual feature is less likely to be the one mentioned. The no-feature state cannot generate this announcement and therefore contributes nothing to either probability.

Because \(\theta\) is uniform on \([0,2\pi)\), the probability of any angular region equals its length divided by \(2\pi\). Over one complete revolution, \(\operatorname{cos}\theta>\frac{1}{2}\) for \(0\le\theta<\frac{\pi}{3}\) and \(\frac{5\pi}{3}<\theta<2\pi\). Each qualifying interval has length \(\frac{\pi}{3}\); their total length is \(\frac{2\pi}{3}\). Therefore

\[P(X>\frac{1}{2})=\frac{(2\pi/3)}{(2\pi)}=\frac{1}{3}.\]

The equality cases \(\theta=\frac{\pi}{3}\) and \(\frac{5\pi}{3}\) have probability zero, so replacing the strict threshold with a non-strict threshold does not change this answer. A tempting calculation, \(\frac{1-1/2}{2}=\frac{1}{4}\), would incorrectly assume X is uniform on \([-1,1]\). It is \(\theta\), not \(\operatorname{cos}\theta\), that was sampled uniformly. The two angular preimages and their lengths are the required cross-check.

Each two-minute count is Poisson with mean \(2\lambda\), so
\[ q(\lambda)=\Pr(\text{empty})=e^{-2\lambda},\qquad 1-q(\lambda)=\Pr(\text{occupied}). \]
The 20 interval indicators are independent. Ignoring the binomial coefficient, which is constant in \(\lambda\), their likelihood is
\[ L(\lambda)=\big(e^{-2\lambda}\big)^8\big(1-e^{-2\lambda}\big)^{12}. \]
Equivalently, regard \(q\) as a Bernoulli empty-interval probability: its likelihood is \(q^8(1-q)^{12}\), maximised at \(\widehat q=8/20=0.4\). Inverting \(q=e^{-2\lambda}\) gives
\[ \boxed{\widehat\lambda=-\tfrac12\log(0.4)=\tfrac12\log(2.5)\approx0.45815\ \text{calls/minute}.} \]
The same result follows by differentiating the log-likelihood: \(\ell'(\lambda)=-16+24e^{-2\lambda}/(1-e^{-2\lambda})\), whose unique interior zero has \(e^{-2\lambda}=0.4\).

The naive calculation \(12/40=0.30\) is occupied intervals per minute, not calls per minute. An occupied interval could contain two, ten, or more calls. At the fitted rate the expected calls in 40 minutes are \(40\widehat\lambda\approx18.33\), which is consistent with observing only 12 occupied intervals.

If all intervals are empty, \(L(\lambda)=e^{-40\lambda}\) is largest at the boundary \(\lambda=0\). If all are occupied, \(L(\lambda)=(1-e^{-2\lambda})^{20}\) increases toward 1 as \(\lambda\to\infty\); there is no finite maximiser. The lack of a finite MLE is a limitation of recording occupancy rather than counts.

Under \(p=1/4\), \(X\sim\operatorname{Binomial}(8,1/4)\). The upper tails are
\[ P_{1/4}(X\ge5)=\frac{1789}{65536}\approx0.027298, \qquad P_{1/4}(X=4)=\frac{5670}{65536}\approx0.086517. \]
Consequently \(P_{1/4}(X\ge4)=7459/65536\approx0.113815>0.05\); no deterministic upper-tail integer cut attains size 5%. The largest permissible nonrandomized rejection region is \(X\ge5\), with actual size 0.027298.

To reach exact size, always reject for \(X\ge5\), never for \(X\le3\), and, only when \(X=4\), draw an independent \(U\sim\operatorname{Uniform}(0,1)\) and reject if \(U<\gamma\). Choose
\[ \gamma=\frac{0.05-1789/65536}{5670/65536} =\frac{7439}{28350}\approx0.262399. \]
This makes the rejection probability at the null boundary exactly \(0.05\). For the composite null \(p\le1/4\), binomial upper tails are increasing in \(p\), so the size is maximized at \(p=1/4\).

At the stated alternative \(p=1/2\), the deterministic rule's power is
\[ P_{1/2}(X\ge5)=\frac{93}{256}=0.36328125, \]
whereas the randomized rule has power
\[ \frac{93}{256}+\gamma\frac{70}{256}\approx0.435031. \]
Randomization is an actual tie-break lottery applied only after exactly four defects are seen, not a fractional defect count. Binomial likelihood ratios increase with \(X\) for any fixed \(p>1/4\), so this boundary-randomized upper-tail test is the most powerful level-0.05 test for such a simple alternative (by the Neyman–Pearson ordering).

The slope fitted by a population regression with an intercept is

\[ \begin{aligned} \beta_{\mathrm{simple}} &=\frac{\operatorname{Cov}(A,Y)}{\operatorname{Var}(A)}\\ &=\frac{12\operatorname{Var}(A)+8\operatorname{Cov}(A,T)+\operatorname{Cov}(A,\varepsilon)}{\operatorname{Var}(A)}\\ &=12+8\frac34=18. \end{aligned} \]

The zero-covariance assumption makes the last error term vanish. The omitted-variable bias is therefore \(18-12=+6\) sales per additional $100 of advertising, a 50% exaggeration relative to the true coefficient. The sign is positive because temperature raises sales and is positively correlated with advertising.

Within the stated structural model, raising advertising by $100 while holding temperature fixed changes expected sales by 12, not 18. A high unconditional fitted association need not justify a budget decision unless the shop controls for temperature or otherwise identifies the causal effect. If the advertising-temperature covariance had the opposite sign, the bias would also switch sign.

Measure each day's return in percentage points, so each \(r_{i}\) is +1 or −1. The eligible reports are \(M_{1}=\frac{r_{1}+r_{2}+r_{3}+r_{4}}{4}\), \(M_{2}=\frac{r_{2}+r_{3}+r_{4}}{3}\) and \(M_{3}=\frac{r_{3}+r_{4}}{2}\). The report displays \(M=\max(M_{1},M_{2},M_{3})\). These sample means are dependent: \(M_{2}\) and \(M_{3}\) share the final two observations, and \(M_{1}\) contains them all.

Enumerate the \(2^{4}=16\) equally likely return paths. The resulting distribution of M, expressed as (value in percentage points : number of paths), is −1:1, −1/2:1, −1/3:1, 0:5, 1/3:2, 1/2:2 and 1:4. The counts sum to 16. Therefore

\[\mathbb{E}[M]=\frac{-1-\frac{1}{2}-\frac{1}{3}+0+\frac{2}{3}+1+4}{16}=\frac{\frac{23}{6}}{16}=\frac{23}{96}\approx0.239583\] percentage points.

For the prespecified window, \(\mathbb{E}[M_{1}]=\frac{\mathbb{E}[r_{1}]+\mathbb{E}[r_{2}]+\mathbb{E}[r_{3}]+\mathbb{E}[r_{4}]}{4}=0\). Thus choosing the start after seeing the complete path adds an expected \(\frac{23}{96}\) percentage points to the displayed per-day average without adding skill or changing any underlying return. The enumeration uses each single path to compute all three windows and therefore incorporates their overlap exactly. Choosing a later start also changes sample length; the answer pertains to the stated average-return reporting rule, not cumulative P&L.