Radix Trading Interview Questions 2026

Since \(N\) is nonnegative and integer-valued, for every outcome

\[ N\ge4\,\mathbf1\{N\ge4\}. \]

Taking expectations yields the mean-based tail bound (Markov's inequality)

\[ 4P(N\ge4)\le E[N],\qquad P(N\ge4)\le\min\left(1,\frac{E[N]}4\right). \]

At \(E[N]=2.4\), the bound is \(2.4/4=\boxed{0.6}\). It is best possible from the mean alone: set \(P(N=4)=0.6\) and \(P(N=0)=0.4\). This distribution has mean \(4(0.6)=2.4\) and needs extra staff with probability exactly 0.6. A lower bound on the staffing frequency beyond zero cannot generally be deduced from the mean without additional constraints on the maximum possible count.

If the mean were instead five, the same inequality gives \(P(N\ge4)\le5/4\), but any probability is already at most one. The useful bound is therefore \(\min(1,5/4)=\boxed1\), and it too is sharp: the deterministic distribution \(N=5\) has mean five and triggers extra staff every day. No assumed Poisson tail or other distributional shape is justified by a mean on its own.

Uniform selection from the resident register makes each resident equally likely, not each household. The four groups contain respectively \(12\times1=12,\ 8\times2=16,\ 6\times4=24\) and \(4\times3=12\) residents. Total population size is \(12+16+24+12=64\).

Service-needing households contain \(16+24=40\) residents, so a random resident's household requires collection with probability \(\frac{40}{64}=\frac{5}{8}=62.5\%\). This is the correct probability for the coordinator's method.

If the coordinator instead drew one of the 30 households uniformly, the service-needing count would be \(8+6=14\) and the probability would be \(\frac{14}{30}=\frac{7}{15}\approx46.67\%\). That answer is wrong for resident-first sampling because a four-person household appears four times in the resident register while a one-person household appears just once.

Independently, a particular household of size s has selection probability \(s/64\) under the actual rule. Summing that weight over the 8 service households of size 2 and the 6 of size 4 yields \(8(2/64)+6(4/64)=40/64=5/8\). The result is a size-biased household probability, not an ordinary fraction of households.

Each independent attempt succeeds with probability \(p=1/5\) and fails with probability \(q=4/5\). Let \(T\) be the number of the first successful attempt, so \(P(T>k)=q^k\). Observing four failures means \(T>4\). Therefore
\[ P(\text{three more failures}\mid T>4) =P(T>7\mid T>4) =\frac{P(T>7)}{P(T>4)} =\frac{q^7}{q^4}=\boxed{\frac{64}{125}}. \]
This is the memoryless property of geometric waiting times: for any nonnegative integers \(s,t\), \(P(T>s+t\mid T>s)=q^t\). It follows from independent Bernoulli trials with a fixed success chance, not from ignoring relevant information.

Let \(R=T-4\) denote the number of additional attempts, including the eventual successful one, conditional on the first four failures. Then \(R\) has the same geometric law as \(T\), with \(P(R=r)=q^{r-1}p\) for \(r\ge1\). Its expected value is
\[ E[R]=\sum_{r\ge1}P(R\ge r)=\sum_{r\ge1}q^{r-1}=\frac1{1-q}=\boxed{5}. \]
Thus \(E[T\mid T>4]=4+E[R]=\boxed{9}\). The next try still succeeds with probability 1/5. If keys were removed permanently after failure or the success rate were unknown and being inferred from failures, the conclusion would require a different model; neither applies here.

For an exponential wait \(T\) with rate \(\lambda\), the survival function is \(\Pr(T\ge t)=e^{-\lambda t}\) for \(t\ge0\). A displayed integer \(Y=k\) describes the whole interval \(k\le T<k+1\), so

\[\Pr(Y=k)=e^{-\lambda k}-e^{-\lambda(k+1)}=e^{-\lambda k}(1-e^{-\lambda}),\quad k=0,1,2,\ldots.\]

These probabilities form a geometric distribution on the nonnegative integers and sum to one: \((1-e^{-\lambda})\sum_{k=0}^{\infty}e^{-\lambda k}=1\). In particular, \(\Pr(Y=0)=1-e^{-\lambda}>0\): a zero display represents a positive-length interval, not a zero-duration point event.

For \(n\) independent readings with total \(S=\sum_i y_i\), the correct likelihood is \(L(\lambda)=(1-e^{-\lambda})^n e^{-\lambda S}\). Hence

\[\ell(\lambda)=n\log(1-e^{-\lambda})-\lambda S,\qquad \ell'(\lambda)=\frac{n}{e^{\lambda}-1}-S.\]

Here \(n=6\) and \(S=0+0+1+2+2+3=8\). Setting the score to zero gives \(e^{\widehat\lambda}=1+n/S=7/4\), so

\[\boxed{\widehat\lambda=\log(7/4)\approx0.5596\ \mathrm{minute}^{-1}}.\]

The log-likelihood is strictly concave for \(\lambda>0\) because \(\ell''(\lambda)=-ne^\lambda/(e^\lambda-1)^2<0\); with \(S>0\), this stationary point is the unique interior global maximum. If instead the integers were wrongly treated as exact continuous exponential observations, the usual rate estimate would be \(\widetilde\lambda=n/\sum_i y_i=6/8=0.75\) per minute. The displayed integers systematically omit fractional waiting time, making their sample mean too small and this rate estimate too large. Algebraically, \(\log(1+x)<x\) for \(x>0\), so the correct \(\log(1+n/S)\) is below the naive \(n/S\) whenever \(S>0\). If every displayed wait had been zero (\(S=0\)), the rounded-data likelihood would instead increase toward its supremum as \(\lambda\to\infty\), with no finite MLE; that edge case does not arise here.

The trial mean is \(\bar X\sim N(\mu,\frac{8^{2}}{64})=N(\mu,1^{2})\), so the standard error is 1 minute. The adoption-relevant hypotheses are \(H_{0}:\mu\le2\) versus \(H_{1}:\mu>2\) minutes. The least favourable point of the null is \(\mu=2\), giving \(z=\frac{3-2}{1}=1\). At one-sided level \(\alpha=0.05\) the critical value is \(z_{0.95}\approx1.64485\), so \(z=1\) does not reject \(H_{0}\). The data do not establish a large enough average saving to cover the fee at the specified level.

The equivalent one-sided 95% lower confidence bound is \(L=\bar X-1.64485(SE)=3-1.64485\approx1.35515\) minutes. Since \(L<2\), the bound does not clear the hurdle. The observed sample mean must meet or exceed \(2+1.64485(1)=3.64485\) minutes to attain a one-sided p-value at most 0.05. The boundary value here uses the usual continuous-normal cutoff, with rounding only for display.

Testing instead \(H_{0}:\mu\le0\) versus \(H_{1}:\mu>0\) yields \(z=\frac{3-0}{1}=3\), which exceeds 1.64485 (one-sided \(p\approx0.00135\)). Thus the same trial gives strong evidence of a positive mean saving but insufficient evidence that the mean exceeds the economically required two minutes. Failing to reject the hurdle null is not proof that \(\mu\le2\); it means adoption has not met the stated evidentiary rule.

Differentiating the ridge objective gives
\[ -2\sum_i x_i(y_i-bx_i)+2\lambda b=0 \quad\Longrightarrow\quad \widehat\beta_\lambda=\frac{\sum_i x_i y_i}{10+\lambda} =\frac{10}{10+\lambda}\widehat\beta_0. \]
Because \(\sum_i x_i y_i=\beta\sum_i x_i^2+\sum_i x_i\varepsilon_i=10\beta+\sum_i x_i\varepsilon_i\), and the errors are independent with variance 40,
\[ E[\widehat\beta_\lambda]=\frac{10\beta}{10+\lambda},\qquad \operatorname{Var}(\widehat\beta_\lambda) =\frac{\sum_i x_i^2(40)}{(10+\lambda)^2} =\frac{400}{(10+\lambda)^2}. \]
At the stipulated true slope \(\beta=3\), the coefficient bias is \(-3\lambda/(10+\lambda)\). For \(x_*=1\), the new response is \(y_*=3+\varepsilon_*\), with fresh error independent of the training data. Its squared prediction error decomposes into irreducible noise, squared coefficient bias and coefficient variance:
\[ M(\lambda)=E[(y_*-\widehat\beta_\lambda)^2] =40+\frac{9\lambda^2}{(10+\lambda)^2} +\frac{400}{(10+\lambda)^2} =\boxed{40+\frac{9\lambda^2+400}{(10+\lambda)^2}}. \]
Ordinary least squares has \(M(0)=40+400/100=44\). Differentiate the reducible part:
\[ \frac{d}{d\lambda}\frac{9\lambda^2+400}{(10+\lambda)^2} =\frac{180\lambda-800}{(10+\lambda)^3}. \]
The derivative changes from negative to positive at \(\boxed{\lambda=40/9}\), the unique minimum for \(\lambda\ge0\). At this point the shrinkage factor is \(10/(10+40/9)=9/13\), and the coefficient-estimation component of risk equals \(9(4)/(9+4)=36/13\). Thus \(M(40/9)=\boxed{40+36/13\approx42.769}\).

To identify where shrinkage helps, compare the reducible part with OLS's value 4:
\[ \frac{9\lambda^2+400}{(10+\lambda)^2}<4 \iff 9\lambda^2+400<4(100+20\lambda+\lambda^2) \iff\lambda(5\lambda-80)<0. \]
For \(\lambda\ge0\), that holds exactly when \(0<\lambda<16\). At 0 and 16 the risks tie; for \(\lambda>16\) excessive shrinkage creates too much bias. The noise floor of 40 is shared by both forecasting methods and cannot be reduced by choosing a different slope penalty.

For an ordered pair \((i,j)\), the four possible assignment states and worker \(i\)'s output are: \((0,0)\mapsto10\), \((0,1)\mapsto12\), \((1,0)\mapsto14\), and \((1,1)\mapsto16\) tasks/day. Under independent assignment, each state has probability \(1/4\). Conditional on \(T_i=1\), the partner is treated with probability \(1/2\), so \(\mathbb E[Y_i\mid T_i=1]=10+4+2(1/2)=15\). Conditional on \(T_i=0\), the partner is still treated with probability \(1/2\), giving \(\mathbb E[Y_i\mid T_i=0]=10+2(1/2)=11\). Hence \(\boxed{\Delta_I=4}\).

Under exactly-one-treated-per-pair assignment, only \((0,1)\) and \((1,0)\) occur, each with probability \(1/2\). A treated worker necessarily has an untreated partner and scores 14; an untreated worker necessarily has a treated partner and scores 12. Consequently \(\boxed{\Delta_{II}=14-12=2}\). The difference is not sampling error: the exclusive design makes the neighbour's treatment negatively related to the worker's own assignment.

For policy evaluation, treating everyone sets \((T_i,T_j)=(1,1)\) and delivers 16 tasks per worker, while treating nobody yields 10. The policy effect is therefore \(\boxed{16-10=6}\) tasks/day.

In general, if output were \(b+\tau T_i+\gamma T_j\), the observed conditional contrast under any assignment design would be

\(\Delta=\tau+\gamma\{\Pr(T_j=1\mid T_i=1)-\Pr(T_j=1\mid T_i=0)\}\).

Independent assignment makes the bracket zero and identifies the direct effect \(\tau=4\) at the design's neighbour-treatment mix. Paired-exclusive assignment makes the bracket \(-1\) and identifies \(\tau-\gamma=2\). The universal policy effect is \(\tau+\gamma=6\) because both own treatment and neighbour treatment change together. Neither design contrast alone answers the all-on-versus-all-off policy question without explicitly accounting for interference.