This one took me a second to set up properly.
First, recognize that each atom independently survives with probability p = e^{-10} after 10 days, so the number alive follows a Binomial(100, p) distribution. Then compute the expected count as 100p, the probability at least one survives as 1 - (1-p)^100, and finally outline a simulation that tracks each atom's state using exponential random variables.
Pro tip: Emphasize the independence assumption and connect the binomial model to practical applications like customer churn or conversion rates, showing you can translate statistical theory into business insights.
Explain that each atom's lifetime is exponential with rate λ = ln(2) per day, so the survival probability after 10 days is p = e^{-10λ} = e^{-10 ln 2} = 2^{-10} = 1/1024. Since atoms decay independently, the number alive follows Binomial(n=100, p=1/1024).
Calculate the expected number alive as E[X] = np = 100/1024 ≈ 0.0977. Compute the probability at least one survives as P(X ≥ 1) = 1 - P(X=0) = 1 - (1-p)^100 ≈ 1 - (1023/1024)^100 ≈ 0.093.
Describe a simulation where for each atom, generate a lifetime from Exponential(λ) and check if it exceeds 10 days. Alternatively, generate a Bernoulli trial with success probability p for each atom. Output the final state (alive/dead) for each atom and optionally the total count.
Mention that the simulation should be repeated many times to verify the distribution, expected value, and survival probability. Discuss how this model applies to independent events in business, such as user retention or conversion.
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The y on x regression is clean: slope is Cov(y,x)/Var(x) = 1/1 = 1, intercept 0.
First, clarify that OLS regression of y on x estimates the linear relationship by minimizing squared errors, and the population slope is Cov(x,y)/Var(x). Then compute the covariance and variances using the given model y = x + epsilon with independent standard normals, and apply the formulas for slope and intercept. Repeat the process for regressing x on y, noting the asymmetry due to error in the regressor.
Pro tip: Emphasize that the slope in regressing x on y is attenuated (biased toward zero) because of measurement error in the regressor, a key insight for causal inference. Also, mention that the intercepts are zero due to zero means, but the slopes differ, illustrating regression to the mean.
State that for regressing y on x, the population slope is β = Cov(x,y)/Var(x) and intercept is α = E[y] - β E[x]. Similarly for regressing x on y, swap roles.
Given x ~ N(0,1), ε ~ N(0,1) independent, y = x + ε. Compute Var(x)=1, Cov(x,y)=Cov(x, x+ε)=Var(x)+Cov(x,ε)=1, and E[x]=E[y]=0. Thus β=1, α=0.
For regressing x on y, slope β' = Cov(x,y)/Var(y). Var(y)=Var(x+ε)=1+1=2. Cov(x,y)=1. So β'=1/2, and intercept α' = E[x] - β' E[y] = 0.
Note that the slopes are not reciprocals (1 vs 0.5) due to error in the regressor. Explain that this demonstrates attenuation bias and the asymmetry of OLS when variables have measurement error.
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Switching wins with probability 2/3, staying wins 1/3.
State the probabilities clearly: staying wins 1/3 of the time, switching wins 2/3. Then justify by explaining that your initial pick has a 1/3 chance of being correct, so the other two doors collectively have a 2/3 chance; when the host reveals a goat, that 2/3 probability collapses onto the remaining unopened door.
Pro tip: Emphasize that the host's action is not random—he knows where the prize is and always reveals a goat—which is why switching doubles your odds. This shows you understand the role of information asymmetry, a key concept in data science.
Clearly say that switching wins with probability 2/3 and staying wins with probability 1/3.
Note that your first pick has a 1/3 chance of being the car, so the other two doors together have a 2/3 chance.
Explain that the host, who knows where the car is, always opens a door with a goat, which does not change the 1/3 probability of your initial pick.
Conclude that the 2/3 probability from the other two doors now rests entirely on the single unopened door, so switching wins 2/3 of the time.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Recognize this as the coupon collector's problem and derive the expected value by summing the expected waiting times for each new face. Use linearity of expectation, where the probability of seeing a new face when k faces remain is k/n, so the expected rolls for that phase is n/k. Sum over k=1 to n to get n * H_n.
Pro tip: Mention that this is a classic problem and that the harmonic number approximation gives n ln n + γn + 1/2, which is useful for large n. Also, note that the variance is also known and can be derived similarly, showing depth.
Recognize that this is the coupon collector's problem: collecting all n distinct faces of a fair die.
Break the process into phases: phase k starts when you have n-k distinct faces and ends when you see a new face, leaving n-k-1 faces unseen.
In phase k, the probability of seeing a new face is k/n, so the expected number of rolls is n/k.
By linearity of expectation, total expected rolls is sum_{k=1}^n n/k = n * H_n, where H_n is the n-th harmonic number.
For large n, H_n ≈ ln n + γ + 1/(2n), so expected rolls ≈ n ln n + γn + 1/2. Mention this to show practical insight.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.