(a) Let X₁, X2,..., Xn be random variables from a normal distribution with unknownmean μ and unknown variance o². We are interested in finding the maximum likelihoodestimates of u and o². Now let û and 2 be the maximum likelihood estimates for uand o2. The probability density function of X, is given by1f(x₁ ; μ, 0²):-22 (Ti-μ)²=2πσ2for - <μ<∞, 0 < o² < ∞ and i = 1, 2, ..., n.Prove thatΣ₁=1 XiΣ=1(xi - μ)2μ =and 2 =nn

Answered: (a) Let X₁, X2,..., Xn be random…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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2)Let X1, X2, ..., Xn be a sample of n units from a population with a probability density function f (x I θ)=θxθ-1 , 0<x<1, θ>0 . According to this: Find the maximum likelihood estimator (MLE) of parameter θ.
Suppose that Y1,Y2,Y3 denote a random sample from an exponential distribution with density function f(y) = Consider the following four estimators of θ: ?1θe−y/θ, y>0,0, otherwise. θˆ =Y, θˆ =Y1+Y2, θˆ =Y1+2Y2, θˆ =Y1+Y2+Y3 =Y ̄. 11223343 Which estimators are unbiased? Among the unbiased estimators, which has the smallest variance?
If the probability density of X is given by f(x) =kx3(1 + 2x)6 for x > 00 elsewhere where k is an appropriate constant, find the probabilitydensity of the random variable Y = 2X 1 + 2X . Identify thedistribution of Y, and thus determine the value of k.
If the random variable T is the time to failure of a commercial product and the values of its probability den-sity and distribution function at time t are f(t) and F(t), then its failure rate at time t is given by f(t)1 − F(t). Thus, thefailure rate at time t is the probability density of failure attime t given that failure does not occur prior to time t.(a) Show that if T has an exponential distribution, thefailure rate is constant. (b) Show that if T has a Weibull distribution (see Exer-cise 23), the failure rate is given by αβt β−1.
Consider a real random variable X with zero mean and variance σ2X . Suppose that wecannot directly observe X, but instead we can observe Yt := X + Wt, t ∈ [0, T ], where T > 0 and{Wt : t ∈ R} is a WSS process with zero mean and correlation function RW , uncorrelated with X.Further suppose that we use the following linear estimator to estimate X based on {Yt : t ∈ [0, T ]}:ˆXT =Z T0h(T − θ)Yθ dθ,i.e., we pass the process {Yt} through a causal LTI filter with impulse response h and sample theoutput at time T . We wish to design h to minimize the mean-squared error of the estimate.a. Use the orthogonality principle to write down a necessary and sufficient condition for theoptimal h. (The condition involves h, T , X, {Yt : t ∈ [0, T ]}, ˆXT , etc.)b. Use part a to derive a condition involving the optimal h that has the following form: for allτ ∈ [0, T ],a =Z T0h(θ)(b + c(τ − θ)) dθ,where a and b are constants and c is some function. (You must find a, b, and c in terms ofthe information…
For a certain psychiatric clinic suppose that the random variable X represents the total time (in minutes) that a typical patient spends in this clinic during a typical visit (where this total time is the sum of the waiting time and the treatment time), and that the random variable Y represents the waiting time (in minutes) that a typical patient spends in the waiting room before starting treatment with a psychiatrist. Further, suppose that X and Y can be assumed to follow the bivariate density function fXY(x,y)=λ2e−λx, 0<y<x, where λ > 0 is a known parameter value. (a) Find the marginal density fX(x) for the total amount of time spent at the clinic. (b) Find the conditional density for waiting time, given the total time. (c) Find P (Y > 20 | X = x), the probability a patient waits more than 20 minutes if their total clinic visit is x minutes. (Hint: you will need to consider two cases, if x < 20 and if x ≥ 20.)
Let X be a Gaussian random variable with zero mean and variance equal to 2. 1-Find the probability density function of the random variableY = 4X + 4 2-Find the probability P(Y>4).
Let the continuous random variable X denote the current measured in a thin copper wire in milliamperes. Assume that the range of X is [4.9, 5.1] mA, and assume that the probability density function of X is f(x) = 5 for 4.9 <= x <= 5.1. What is the variance?
X is an exponential random variable with λ =1 and Y is a uniform random variable defined on (0, 2). If X and Y are independent, find the PDF of Z = X-Y2
If the independent random variables X and Y havethe marginal densitiesf(x) =⎧⎪⎪⎨⎪⎪⎩12for 0 < x < 20 elsewhereπ(y) =⎧⎪⎪⎨⎪⎪⎩13for 0 < y < 30 elsewherefind(a) the joint probability density of X and Y;(b) the value of P(X2 + Y2 > 1).
Two random variables X and Y have a joint cumulative distribution function given by FXY(x, y) = 1/2 [u(x-2) + u(x-3)] {(1 – exp(-y/2)) u(y), then the marginal probability density function fx(x) is given by
X1 and X2 are two discrete random variables, while the X1 random variable takes the values x1 = 1, x1 = 2 and x1 = 3, while the X2 random variable takes the values x2 = 10, x2 = 20 and x2 = 30. The combined probability mass function of the random variables X1 and X2 (pX1, X2 (x1, x2)) is given in the table below a) Find the marginal probability mass function (pX1 (X1)) of the random variable X1.b) Find the marginal probability mass function (pX2 (X2)) of the random variable X2.c) Find the expected value of the random variable X1.d) Find the expected value of the random variable X2.e) Find the variance of the random variable X1.f) Find the variance of the random variable X2.g) pX1 | X2 (x1 | x2 = 10) Find the mass function of the given conditional probability.h) pX2 | X1 (x2 | x1 = 2) Find the mass function of the given conditional probability.i) Are the random variables X1 and X2 independent? Show it. The combined probability mass function of the random variables X1 and X2 is below

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