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Question 2. [18 marks] Consider the two-dimensional random vector (X, Y), where X is a continuous random variable and Y is a scalar binary-valued random variable. Assume the following conditional distribution of X given Y: X|Y = i ~ N(?_i, ?_i^2), i = 0, 1, (2) where the means and covariances are distinct for all i = 0, 1. The marginal distribution of Y is P(Y = 0) = ? and P(Y = 1) = 1 - ? for some ? ? [0, 1]. (a). [6 marks ] Suppose we observe n i.i.d sample observations of this form, {(X_j, Y_j)}^n_{j=1}. Find the minimal sufficient statistics for unknown parameters {(?_i, ?_i^2)}_{i=0,1} and ?. (b). [6 marks ] Derive an expression for the marginal distribution of X. The marginal distribution is also a function of {(?_i, ?_i^2)}_{i=0,1} and ?. (c). [6 marks ] Suppose we can only observe X and not Y, and that we have observations {X_j}^n_{j=1} generated according to the marginal distribution derived in part (b). What are the minimal sufficient statistics for {(?_i, ?_i^2)}_{i=0,1} and ? in this situation?

          Question 2. [18 marks] Consider the two-dimensional random vector (X, Y), where X is a continuous random variable and Y is a scalar binary-valued random variable. Assume the following conditional distribution of X given Y:

X|Y = i ~ N(?_i, ?_i^2), i = 0, 1, (2)

where the means and covariances are distinct for all i = 0, 1. The marginal distribution of Y is P(Y = 0) = ? and P(Y = 1) = 1 - ? for some ? ? [0, 1].

(a). [6 marks ] Suppose we observe n i.i.d sample observations of this form, {(X_j, Y_j)}^n_{j=1}.
Find the minimal sufficient statistics for unknown parameters {(?_i, ?_i^2)}_{i=0,1} and ?.

(b). [6 marks ] Derive an expression for the marginal distribution of X. The marginal
distribution is also a function of {(?_i, ?_i^2)}_{i=0,1} and ?.

(c). [6 marks ] Suppose we can only observe X and not Y, and that we have observations
{X_j}^n_{j=1} generated according to the marginal distribution derived in part (b). What
are the minimal sufficient statistics for {(?_i, ?_i^2)}_{i=0,1} and ? in this situation?

Show more…

Question 2. [18 marks] Consider the two-dimensional random vector (X, Y), where X is a continuous random variable and Y is a scalar binary-valued random variable. Assume the following conditional distribution of X given Y:

X|Y = i N(?i, ?i^2), i = 0, 1, (2)

where the means and covariances are distinct for all i = 0, 1. The marginal distribution of Y is P(Y = 0) = ? and P(Y = 1) = 1 - ? for some ? ? [0, 1].

(a). [6 marks ] Suppose we observe n i.i.d sample observations of this form, (Xj, Yj)^nj=1.
Find the minimal sufficient statistics for unknown parameters (?i, ?i^2)i=0,1 and ?.

(b). [6 marks ] Derive an expression for the marginal distribution of X. The marginal
distribution is also a function of (?i, ?i^2)i=0,1 and ?.

(c). [6 marks ] Suppose we can only observe X and not Y, and that we have observations
Xj^nj=1 generated according to the marginal distribution derived in part (b). What
are the minimal sufficient statistics for (?i, ?i^2)i=0,1 and ? in this situation?

Added by Wanda F.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Andreas Papavassiliou

11/15/2024

Step 1

The minimal sufficient statistics for the unknown parameters are the sample means and variances for each group, as well as the proportion of observations with $Y=0$: $$ T(\mathbf{x}, \mathbf{y}) = \left( \frac{1}{n_0} \sum_{i=1}^{n_0} x_i, \frac{1}{n_1} Show more…

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Question 2. [18 marks] Consider the two-dimensional random vector (X, Y), where X is a continuous random variable and Y is a scalar binary-valued random variable. Assume the following conditional distribution of X given Y: X|Y = i ~ N(μ_i, σ_i^2), i = 0, 1, where the means and covariances are distinct for all i = 0, 1. The marginal distribution of Y is P(Y = 0) = θ and P(Y = 1) = 1 - θ for some θ ∈ [0, 1]. (a). [6 marks] Suppose we observe n i.i.d sample observations of this form, {(X_j, Y_j)}^n_{j=1}. Find the minimal sufficient statistics for unknown parameters {(μ_i, σ_i^2)}_{i=0,1} and θ. (b). [6 marks] Derive an expression for the marginal distribution of X. The marginal distribution is also a function of {(μ_i, σ_i^2)}_{i=0,1} and θ. (c). [6 marks] Suppose we can only observe X and not Y, and that we have observations {X_j}^n_{j=1} generated according to the marginal distribution derived in part (b). What are the minimal sufficient statistics for {(μ_i, σ_i^2)}_{i=0,1} and θ in this situation?

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00:06 So suppose we observe sample observations of the form xj, y, j, spire of n, j is equal to 1, from j is equal to 1 to n.

00:24 Let's find the minimal sufficient statistics for the oner of parameters, mu, mean value, i, and variance i is equals 0 and 1.

00:38 In theta, we'll use the factorization theorem, which states that statistic txy is sufficient, for parameter theta if the joint distribution of xy can be factored as the following for parameters g times h so we write the likelihood function so the joint probability density function pdf of x j and y j given our parameters and theta is the following so you have m not um sorry new not new one sigma not squared sigma 1 squared and theta so now this is basically mu 0 sigma not 0 squared you 1 squared probability of y j times f of x j given y j so for yj so for yj equal to 0 0 we have probability yj close to zero theta and xj is equal to 0 is equal to 0 because it follows a conditional distribution and therefore it would be equal to theta over square of 2 pi s not squared times the exponential of minus xj minus m not squared so sigma not squared and then for yj is equal to one probability yj is equal to one is one minus theta x j ij equal to one therefore xj yj is equal to one speed 1 minus theta over square root of 2 pi sigma 1 squared as exponential of minus xj minus new 1 squared or 2 sigma 1 squared and so we combine both cases and we see that the likelihood for all observations, just the product is the product from j is equal to 1 to n of this, it's the power of 1 minus yj times the power of yj.

06:14 So if we simplify the likelihood, we get theta, it's the power of n -not.

06:35 Data 1.

06:36 Sorry.

06:44 We get 1 minus theta...

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