The sample mean is the obvious answer and I got there fine.
Start by recognizing that with i.i.d. unbiased measurements and known variance, the sample mean is the natural estimator. Derive its variance as σ²/n, and briefly justify why it's optimal (e.g., by the Gauss-Markov theorem or Cramér-Rao lower bound).
Pro tip: Mention that the sample mean is the maximum likelihood estimator under normality and is also the best linear unbiased estimator (BLUE) even without normality. This shows depth and connects to broader statistical principles.
State that measurements X₁, X₂, ..., Xₙ are i.i.d. with mean μ (true temperature) and known variance σ². Assume unbiasedness: E[Xᵢ] = μ.
Propose the sample mean X̄ = (1/n) Σ Xᵢ as the estimator for μ. Explain that it's intuitive and leverages all data equally.
Compute Var(X̄) = Var((1/n) Σ Xᵢ) = (1/n²) Σ Var(Xᵢ) = (1/n²)(nσ²) = σ²/n. Show the steps clearly.
Mention that X̄ is the minimum variance unbiased estimator (MVUE) for μ under normality, and by the Gauss-Markov theorem it's BLUE. Also note it achieves the Cramér-Rao lower bound.
Summarize that the estimator is unbiased with variance decreasing as 1/n, so more measurements increase precision. Optionally, mention confidence intervals.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This is the weighted least squares / inverse-variance weighting setup.
Start by setting up the problem: you have two unbiased estimators with variances σ₁² and σ₂². The goal is to find weights w and 1-w that minimize the variance of the combined estimator while maintaining unbiasedness. Use the method of Lagrange multipliers or simply minimize the variance function with respect to w, then derive the optimal weights and the resulting variance.
Pro tip: Emphasize that the optimal weights are inversely proportional to the variances, which is a fundamental result in estimation theory. Also, mention that this is a special case of the Gauss-Markov theorem for combining independent unbiased estimators.
Let the combined estimator be θ̂ = w θ̂₁ + (1-w) θ̂₂, where θ̂₁ and θ̂₂ are the two unbiased estimators. Since both are unbiased, any convex combination is also unbiased.
Since the measurements are independent, the variance of the combined estimator is Var(θ̂) = w² σ₁² + (1-w)² σ₂².
Take the derivative of Var(θ̂) with respect to w, set it to zero, and solve for w. This yields w* = σ₂² / (σ₁² + σ₂²).
The optimal weight for the first estimator is w* = σ₂² / (σ₁² + σ₂²), and for the second estimator is 1 - w* = σ₁² / (σ₁² + σ₂²). These are inversely proportional to their variances.
Substitute w* back into the variance expression to get Var(θ̂*) = (σ₁² σ₂²) / (σ₁² + σ₂²). This is the minimum achievable variance.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, define the random walk mathematically: let X_n and Y_n be the net horizontal and vertical displacements after n steps. Then compute Cov(X_n, Y_n) and Var(X_n) to find the correlation. Use symmetry and independence of step directions to simplify the calculation.
Pro tip: Emphasize that the correlation is zero because the x and y coordinates are uncorrelated due to the independence of horizontal and vertical movements in each step. This demonstrates a deep understanding of the underlying stochastic process.
Let X_n and Y_n represent the net displacement in the x and y directions after n steps. Each step is a random vector (ΔX, ΔY) taking values (1,0), (-1,0), (0,1), (0,-1) with equal probability 1/4.
Calculate E[X_n] = 0, E[Y_n] = 0 by symmetry. Compute Var(X_n) = E[X_n^2] = n/2 and similarly Var(Y_n) = n/2, since each step contributes 1 to either x or y with probability 1/2.
Find Cov(X_n, Y_n) = E[X_n Y_n] - E[X_n]E[Y_n]. Since E[X_n]=E[Y_n]=0, Cov = E[X_n Y_n]. Show E[X_n Y_n] = 0 by noting that at each step, ΔX and ΔY cannot both be nonzero, and their product is always 0.
Use the formula Corr(X_n, Y_n) = Cov(X_n, Y_n) / sqrt(Var(X_n) Var(Y_n)). Since Cov = 0, the correlation is 0.
Explain that the x and y coordinates are uncorrelated, meaning knowledge of one coordinate provides no linear information about the other. This is intuitive because horizontal and vertical movements are independent in each step.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, clarify the model: a 2D simple symmetric random walk where each step is ±1 in x or y with equal probability. Then compute the correlation between |X_n| and |Y_n| by analyzing their joint distribution or using properties of simple random walks, and if a closed form is elusive, determine the sign and asymptotic behavior via simulation or known results.
Pro tip: Emphasize that while X_n and Y_n are independent, their absolute values are not; the correlation is positive due to the shared constraint that the total number of steps is fixed. This subtlety often trips up candidates.
State the assumptions: each step is either ±1 in x or y with equal probability, and steps are independent. Confirm that X_n and Y_n are independent random variables.
Write the correlation formula: Corr(|X_n|, |Y_n|) = (E[|X_n||Y_n|] - E[|X_n|]E[|Y_n|]) / (SD(|X_n|) SD(|Y_n|)). Note that E[|X_n||Y_n|] = E[|X_n|]E[|Y_n|] if |X_n| and |Y_n| were independent, but they are not.
Recognize that |X_n| and |Y_n| are negatively correlated in terms of the number of steps in each direction? Actually, they are positively correlated because if |X_n| is large, it likely means many steps were in the x-direction, leaving fewer steps for the y-direction, which tends to make |Y_n| smaller. So the correlation is negative? Wait, need to check: If many steps in x, then |X_n| large, but then fewer steps in y, so |Y_n| tends to be smaller. So negative correlation. But simulation shows positive? Let's think: For n=2, possible outcomes: (2,0), (0,2), (1,1), (1,-1), (-1,1), (-1,-1), (0,0)? Actually, steps: each step chooses x or y with prob 1/2, and sign ±1. So after n steps, the number of x-steps K ~ Binomial(n, 1/2). Given K, X_n is sum of K ±1's, so |X_n| has distribution related to K. Similarly, Y_n has n-K steps. So |X_n| and |Y_n| are dependent through K. If K is large, |X_n| tends to be larger, and n-K small, so |Y_n| tends to be smaller. So negative correlation. But wait, is that true? Let's test n=1: possible outcomes: (1,0), (-1,0), (0,1), (0,-1). |X| and |Y|: (1,0), (1,0), (0,1), (0,1). So pairs: (1,0) twice, (0,1) twice. E[|X|]=0.5, E[|Y|]=0.5. E[|X||Y|]=0. Cov = -0.25. Correlation = -0.25 / (0.5*0.5) = -1. So negative correlation. For n=2: possible outcomes: (2,0): |X|=2, |Y|=0; (0,2): 0,2; (1,1): 1,1; (1,-1): 1,1; (-1,1): 1,1; (-1,-1): 1,1; (0,0): 0,0? Actually, (0,0) can occur if one step x and one step y with opposite signs? No, (0,0) means X=0 and Y=0, which requires equal number of + and - in both, but total steps 2, so impossible unless one step x and one step y with signs? If one step x=+1 and one step y=-1, then X=1, Y=-1, not (0,0). So (0,0) not possible. So outcomes: (2,0) prob 1/8? Let's compute: each step: choose axis (1/2) and sign (1/2). So probability of (2,0): both steps x and both +: (1/2 * 1/2)^2? Actually, each step: P(x,+)=1/4, x,-=1/4, y,+ =1/4, y,-=1/4. For (2,0): need two x,+ steps: prob (1/4)^2 = 1/16. Similarly (2,0) can also be two x,- steps? That gives (-2,0), so |X|=2, |Y|=0. So (2,0) and (-2,0) each prob 1/16, total 1/8 for |X|=2, |Y|=0. Similarly (0,2) and (0,-2) total 1/8. Now (1,1): need one x step and one y step, with signs such that X=1, Y=1. Possibilities: x+ then y+ (prob 1/16), y+ then x+ (1/16) total 1/8. Similarly (1,-1): x+ then y- (1/16), y- then x+ (1/16) total 1/8. (-1,1): x- then y+ (1/16), y+ then x- (1/16) total 1/8. (-1,-1): x- then y- (1/16), y- then x- (1/16) total 1/8. So all four combinations of |X|=1, |Y|=1 have total probability 1/2. Also (0,0) not possible. So distribution: P(|X|=2, |Y|=0)=1/8, P(|X|=0, |Y|=2)=1/8, P(|X|=1, |Y|=1)=1/2. Check sum: 1/8+1/8+1/2=1. Now compute E[|X|] = 2*(1/8) + 0*(1/8) + 1*(1/2) = 1/4 + 1/2 = 3/4. Similarly E[|Y|]=3/4. E[|X||Y|] = (2*0)*(1/8) + (0*2)*(1/8) + (1*1)*(1/2) = 1/2. Cov = 1/2 - (3/4)^2 = 1/2 - 9/16 = -1/16. Var(|X|) = E[|X|^2] - (E[|X|])^2. E[|X|^2] = 4*(1/8) + 0*(1/8) + 1*(1/2) = 1/2 + 1/2 = 1. So Var = 1 - 9/16 = 7/16. SD = sqrt(7/16) ≈ 0.6614. Correlation = (-1/16) / (7/16) = -1/7 ≈ -0.1429. So negative correlation. So indeed negative. So the correlation is negative. Asymptotically, what happens? For large n, X_n and Y_n are approximately independent? But |X_n| and |Y_n| are dependent through the number of steps. Actually, as n grows, the number of x-steps K ~ Binomial(n, 1/2) has mean n/2 and variance n/4. The fluctuations in K are O(sqrt(n)). Given K, X_n is sum of K ±1's, so |X_n| ~ sqrt(K) * something? Actually, X_n is approximately normal with variance K, so |X_n| has expectation ~ sqrt(2K/π). Similarly |Y_n| ~ sqrt(2(n-K)/π). So the correlation between |X_n| and |Y_n| arises from the negative correlation between K and n-K. Since K and n-K are perfectly negatively correlated (K + (n-K) = n), the correlation between sqrt(K) and sqrt(n-K) is negative. As n grows, the relative fluctuations in K become smaller, so the correlation might vanish? Let's compute asymptotic correlation. We can use delta method. Let K = n/2 + sqrt(n)/2 * Z, where Z ~ N(0,1). Then |X_n| ≈ sqrt(2K/π) = sqrt(2(n/2 + sqrt(n)/2 Z)/π) = sqrt(n/π) * sqrt(1 + Z/sqrt(n)) ≈ sqrt(n/π) (1 + Z/(2 sqrt(n))). Similarly |Y_n| ≈
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.