This felt like a warmup but I fumbled the cdf notation at first, wrote it without the standardization step and had to backtrack.
Start by writing the PDF and CDF of the Normal distribution with mean μ and variance σ², then standardize the random variable X₁ to Z = (X₁ - μ)/σ. Use the properties of the standard Normal distribution to express the probability in terms of Φ, recognizing that P(μ - σ ≤ X₁ ≤ μ + σ) = P(-1 ≤ Z ≤ 1) = Φ(1) - Φ(-1) = 2Φ(1) - 1.
Pro tip: Emphasize that the result is a constant (approximately 0.6826) independent of μ and σ, which is the well-known 68-95-99.7 rule. This shows you understand the practical significance of the standard deviation in a Normal distribution.
State the probability density function of the Normal distribution: f(x) = (1/(σ√(2π))) * exp(-(x-μ)²/(2σ²)) for x ∈ ℝ.
State the cumulative distribution function: F(x) = P(X ≤ x) = ∫_{-∞}^{x} f(t) dt, which has no closed form in elementary functions.
Define Z = (X₁ - μ)/σ, which follows a standard Normal distribution with mean 0 and variance 1. Then rewrite the probability: P(μ - σ ≤ X₁ ≤ μ + σ) = P(-1 ≤ Z ≤ 1).
Use the standard Normal CDF Φ(z) = P(Z ≤ z) to write P(-1 ≤ Z ≤ 1) = Φ(1) - Φ(-1).
Apply the symmetry property Φ(-z) = 1 - Φ(z) to get Φ(1) - (1 - Φ(1)) = 2Φ(1) - 1.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
The first two cases are textbook and I got through them fine.
Start by writing the likelihood function for a normal sample, then derive the MLEs for each scenario by maximizing the log-likelihood. For the constrained case, use the fact that the parameter space is restricted, so the MLE is the projection of the unconstrained MLE onto the constraint set. Explain the boundary solution when the unconstrained estimate is negative.
Pro tip: Emphasize that the constrained MLE is not simply setting the negative estimate to zero; it's the projection onto the feasible region, which for a single constraint is indeed zero, but you should justify it via the Karush-Kuhn-Tucker conditions or by checking the likelihood at the boundary.
Write the joint probability density function for i.i.d. normal observations, then take the natural logarithm to obtain the log-likelihood function.
Maximize the log-likelihood with respect to μ only, treating σ² as fixed. The MLE is the sample mean.
Maximize the log-likelihood jointly over μ and σ². The MLEs are the sample mean for μ and the biased sample variance (dividing by n) for σ².
Maximize the log-likelihood subject to μ ≥ 0. The unconstrained MLE is the sample mean; if it is non-negative, it remains the MLE. If it is negative, the constrained MLE is 0, and σ² MLE is adjusted accordingly.
When the unconstrained estimate is negative, the likelihood is decreasing in μ over the feasible region, so the maximum occurs at the boundary μ = 0. The MLE for σ² becomes the second moment about zero (i.e., the mean of squared observations).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by stating the bivariate normal setup and the known result that the conditional distribution of X given Y=y is normal. Then derive the conditional mean and variance using the joint parameters, write the pdf and cdf, and finally compute the tail probability using the standard normal CDF Φ. Emphasize the linearity and homoscedasticity of the conditional distribution.
Pro tip: Mention that the conditional variance does not depend on y, which is a unique property of the normal distribution and important for modeling. Also, connect this to practical applications like Gaussian processes or linear regression.
Define the joint distribution with means μ_X, μ_Y, variances σ_X^2, σ_Y^2, and correlation ρ. Write the joint pdf or simply state the parameters.
Use the known formulas: E[X|Y=y] = μ_X + ρ (σ_X/σ_Y)(y - μ_Y) and Var(X|Y=y) = σ_X^2 (1 - ρ^2). Explain that these come from the properties of the multivariate normal.
State that X|Y=y ~ N(μ_{X|Y}, σ_{X|Y}^2). Write the pdf explicitly and express the cdf in terms of Φ: F_{X|Y}(x|y) = Φ((x - μ_{X|Y})/σ_{X|Y}).
For P(X > t | Y=y), standardize: P(X > t | Y=y) = 1 - Φ((t - μ_{X|Y})/σ_{X|Y}) = Φ((μ_{X|Y} - t)/σ_{X|Y}) due to symmetry.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
No, the MLE divides by n and has expected value (n−1)/n times σ², so it's biased downward.
Start by stating the standard MLE of σ² is biased, then derive the bias and present the unbiased estimator. Explain the relationship between the two estimators and discuss practical implications.
Pro tip: Mention that while the unbiased estimator is often preferred, the MLE's bias decreases with sample size, and in some contexts (e.g., predictive modeling) the biased version might be used for its lower MSE.
State that the MLE of σ² is the average of squared residuals: σ̂²_MLE = (1/n) Σ (X_i - X̄)².
Compute the expectation: E[σ̂²_MLE] = ((n-1)/n) σ², which is less than σ², hence biased.
Introduce the unbiased estimator: s² = (1/(n-1)) Σ (X_i - X̄)², which corrects the bias.
Show that s² = (n/(n-1)) σ̂²_MLE, and discuss how the bias diminishes as n grows.
Mention that the choice depends on context: unbiasedness for inference, MLE for large samples or when minimizing MSE is desired.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.