概率作业代写 Probability代写 Sampling Distributions代写

Assignment

概率作业代写 1 1.3 The 95% confidence interval for θ means that the true value θ shall be inside the confidence interval with a probabilty of 95%. 1.4 Sometimes

1

1.3

The 95% confidence interval for θ means that the true value θ shall be inside the confidence interval with a probabilty of 95%.

1.4

Sometimes the UMPT doesn’t exist because of the requirement for simple hypothesis.

1.5

Yes, UMVUE is unique.

1.6

The estimator is a function of input data that returns an estimation of a parameter by maximizing the likelihood of data observed. The estimate is an estimation of a parameter, based on a realization of the data ( as in point estimate )

1.7

1. Cumulative distribution function technique

2. Transformation technique

3. Moment generating function technique

1.8

Generally there will be a trade-off. To minimize one type of error will tend to increase the probability of another type of error.

1.10

For MPT, both H₀ and H_A are simple hypotheses. For UMPT, they are composite hypotheses.

2 概率作业代写

So we have:

E(Y ) = M’_Y (0) = E(NX × e^tNX) (10)

= E(N)E(X × e₀ ) (11)

= E(N) × M’_X(0) (12)

And by the fact that N and X are independent: E(f(N)g(X)) = E(f(N))E(g(X))

σ²(Y ) = M’’_Y (0) − [M’_Y (0)]²  (13)

= E(N²X² × e⁰) − E²(NX × e⁰ ) (14)

= E(N²)E(X²) − E²(NX) (15)

= E(N² )(E(X²) − E²(X)) + E²(X)E(N²) − E²(X)E²(N) (16)

= E(N)(E(X²) − E²(X)) + (E(N²) − E²(N))E²(X) (17)

= E(N)V ar(X) + V ar(N)E²(X) (18)

3 概率作业代写

3.1 part a

To find the moment estimator for θ, note that

3.4 part d

Given the prior distribution of θ ∼ U(0, 1), the density function for θ in (0,1) is π(θ) = 1. The posterior distribution of θ is:

π(θ; x₁, . . . , x_n) = π(θ)L(θ; x₁, . . . , x_n) (33)

= 1 ∗ θⁿ (1 − θ)^nM1(x)−n              (34)

Note that this is the same as part (c) because the prior of U(0,1) is uninformative. So the Bayesian estimator of θ with this uniform prior is still the same as the MLE of θ:

4 概率作业代写

4.1 part a

By the Neyman-Pearson Lemma, the likelihood ratio of observing data x₁, . . . , x_n under H₀ and H_A  is:

5 概率作业代写

5.1 part a

5.2 part b

When H₀ : Exp(0.005) is true, the expected number of lightbulb with each lifespan that fall in the intervals are shown in the table below:

Time of life	¡ 100	[100, 200)	[200, 300)	≥ 300
Number of bulbs	121	78	43	58
Expected number	118.04	71.60	43.42	66.94

The goodness of fit test is:

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