Concept explainers
This exercise demonstrates that, in general, the results provided by Tchebysheff’s theorem cannot be improved upon. Let Y be a random variable such that
a Show that E(Y) = 0 and V(Y) = 1/9.
b Use the
*c In part (b) we guaranteed E(Y) = 0 by placing all probability mass on the values –1, 0, and 1, with p(–1) = p(1). The variance was controlled by the probabilities assigned to p(–1) and p(1). Using this same basic idea, construct a probability distribution for a random variable X that will yield
*d If any k > 1 is specified, how can a random variable W be constructed so that
a
Prove that
Explanation of Solution
Calculation:
The expected value of a random variable is the sum of the product of each observation with corresponding probability.
Hence, the expected value is obtained as follows:
Similarly the variance of a random variable is the expected value of the squared deviation from the mean.
Hence, it is proved that
b
Find the value of
Compare the exact probability with the upper bound of the Tchebysheff’s theorem.
Answer to Problem 169E
The value of
Explanation of Solution
Calculation:
Tchebysheff’s Theorem:
Let Y be a random variable with mean
Thus, using the Tchebysheff’s theorem it can be obtained that,
Similarly, using the given information the required probability is obtained as follows:
Hence, the value of
Thus, it can be said the exact probability is same with the upper bound of the Tchebysheff’s theorem when
c.
Generate a probability distribution for a random variable X that will yield
Answer to Problem 169E
The probability distribution of another sufficient random variable is,
Explanation of Solution
Calculation:
Consider the random variable X which is symmetric around mean 0.
The probability distribution of X is,
It is needed to find the value of p.
The mean of X is,
The variance of X is,
Thus, using Tchebysheff’s theorem it can be said that,
Now, it is obvious that
Hence,
Thus, the probability distribution of another sufficient random variable is,
d.
Explain the procedure of obtaining a random variable W such that
Explanation of Solution
Calculation:
Consider the random variable W which is symmetric around mean 0.
The probability distribution of W is,
It is needed to find the value of p.
The mean of W is,
The variance of W is,
Thus, using Tchebysheff’s theorem it can be said that,
It is clear that W can take only one positive value. Hence, if
Hence,
Thus, the probability distribution of another sufficient random variable is,
Want to see more full solutions like this?
Chapter 3 Solutions
EBK MATHEMATICAL STATISTICS WITH APPLIC
- 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…arrow_forwardIf X is a continuous random variable with X ∼ Uniform([0, 2]), what is E[X^3]?arrow_forwardLet X be a Poisson random variable with E(X) = 3. Find P(2 < x < 4).arrow_forward
- consider x-U(0,1). Let Y=eX and Z=X2. Find the expected value of random variable T=Y+Zarrow_forwardSuppose that the random variables Y1 and Y2 have joint probability distribution function. f(y1, y2) = 2, 0 ≤ y1 ≤ 1, 0 ≤ y2 ≤ 1, 0 ≤ y1 + y2 ≤ 1, 0, elsewhere (a) Use R to calculate P(Y1 ≥ 1⁄6 | Y2 ≤ 1⁄5). (Round your answer to four decimal places.) P(Y1 ≥ 1⁄6 | Y2 ≤ 1⁄5) = (b) Use R to calculate P(Y1 ≥ 1⁄6 | Y2 = 1⁄5). (Round your answer to four decimal places.) P(Y1 ≥ 1⁄6 | Y2 = 1⁄5) =arrow_forwardIf X and Y are Gaussian random variables then what is E[XY]?arrow_forward
- For any continuous random variables X, Y , Z and any constants a, b, show the following from the definition of the covariance:arrow_forwardFor two random variables X and Y where X∼ exponential(0.5) and Y|X = x ∼ N(5,x2), find E(E(X|Y ))arrow_forwardConsider a random variable Y with PDF Pr(Y=k)=pq^(k-1),k=1,2,3,4,5....compute for E(2Y)arrow_forward
- MATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th...StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C...StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. WallnauPublisher:Cengage Learning
- Elementary Statistics: Picturing the World (7th E...StatisticsISBN:9780134683416Author:Ron Larson, Betsy FarberPublisher:PEARSONThe Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. Moore, William I. Notz, Michael A. FlignerPublisher:W. H. FreemanIntroduction to the Practice of StatisticsStatisticsISBN:9781319013387Author:David S. Moore, George P. McCabe, Bruce A. CraigPublisher:W. H. Freeman