Due date: Tuesday 4/28, 11:59pm

Mean square error

01

05

MMSE linear estimator from joint PMF

Suppose X and Y have the following joint PMF:

YX101
1161120
318112124
512411218
7011216

(a) Find the minimal MSE linear estimator for X in terms of Y.

(b) What is the MMSE error for this linear estimator?

(c) Use (a) to estimate X given Y=1 and Y=5.

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02

06

MMSE linear estimator from joint density

Consider this joint PDF:

fX,Y(x,y)={32(x2+y2)x,y[0,1]0 otherwise 

(a) What is the minimal MSE linear estimator for X in terms of Y?

(b) What is the linear estimate of X given Y=0.7?

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03

07

Telemetry signal

A telemetry signal, T, transmitted from a temperature sensor on a communications satellite is a Gaussian random variable with E[T]=0 and Var[T]=9. The receiver at mission control receives R=T+X, where X is a noise voltage independent of T with PDF:

fX(x)={1/63x30 otherwise 

The receiver uses R to calculate a linear estimate of the telemetry voltage:

T^L(R)=aR+b

(a) What is E[R], the expected value of the received voltage?

(b) What is Var[R], the variance of the received voltage?

(c) What is Cov[T,R], the covariance of the transmitted voltage and the received voltage?

(d) What is the correlation coefficient ρT,R of T and R?

(e) What are a and b, the optimum mean square values of a and b in the linear estimator?

(f) What is eL, the minimum mean square error of the linear estimate?

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Review

04

06

Bits received in error

In a digital communication channel, it is assumed that a bit is received in error with probability 4.3×105. Someone challenges this hypothesis: they believe the error rate is higher than 4.3×105. Assume 100,000 bits are transmitted. Design a one-tailed significance test using α=0.05 and N, the number of bits received in error, to decide whether to reject the hypothesis that the error rate is 4.3×105. Your rejection region should be of the form {Nc}. You do not have to use the continuity correction.

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05

05

CAT scan for tumors

When a brain is scanned in a CAT scan, analysis of the results yields a rating of 1, 2, 3, or 4. This represents (imperfect) evidence of whether there is a tumor.

X1234
No tumor:PX(k)0.40.30.20.1
X1234
With tumor:PX(k)0.00.10.30.6

Suppose that, of people who get CAT scans, 20% do have a tumor.

Furthermore, assume that declaring there is no tumor when there is one is ten times worse than declaring there is a tumor when there isn’t one.

Design an MC test to determine which ratings should be classified as tumors.

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