Let X be continuous with pdf f(x)=e* if 0 < x < ∞, and zero otherwise: . Use Chebychev's inequality to obtain a lower bound on P(-1.5
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Rate of Change
The relation between two quantities which displays how much greater one quantity is than another is called ratio.
Slope
The change in the vertical distances is known as the rise and the change in the horizontal distances is known as the run. So, the rise divided by run is nothing but a slope value. It is calculated with simple algebraic equations as:
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- The PDF of a continuous random variable Y is provided as the image shown below. Show the CDF of G_y (Y). Do not write your final answer in terms of the constant c. Provide a graph for the CDFf (x) = 4xc for 0 ≤ x ≤ 1 ; Find the constant c so that f (x) is apdf of some random variable X, and then find the cdf, F (x) = P(X ≤ x). Sketch graphsof the pdf f (x) and the cdf F (x), and find the mean μ and variance σ2.An individual has a vNM utility function over money of u(x) = Vx, where x is final wealth. Assume the individual currently has $16. He is offered a lottery with three possible outcomes; he could gain an extra $9, lose $7, or not lose or gain anything. There is a 15% probability that he will win the extra $9. What probability, p, of losing $7 would make the individual indifferent between to play and to not play the lottery? (Make sure to answer in the form, 0.X, i.e. 0.25) Enter your answer here
- Which statements are true? Select one or more: a. Markov’s inequality is only useful if I am interested in that X is larger than its expectation. b. Chebyshev’s inequality gives better bounds than Markov’s inequality. c. Markov’s inequality is easier to use. d. One can prove Chebyshev’s inequality using Markov’s inequality with (X−E(X))2.Let a random variable X has the following function:FX(x) = { 0 ; x < 0(1/2)+(x/2) ; 0 ≤ x < 11 ; x ≥ 1. Verify that FX(x) is a CDF. Determine (i) the PDF of X, (ii) E[e^X], (iii) P[X = 0|X≤ 0.5].3. The expectation operator E can be applied to a random vector. Specif- ically, if X = [X1 X2 …… Xn]", then EX = [EX1 EX2 .… EXn]". Show that E[(X – EX)(X – EX)"]= [cov(X¡, X;)]"j=1 •
- Example 2.1.1 Let S = R, and suppose P is a probability measure on R. Define F(x) by F(x) = P((-∞, x]), XER. (2.3) Then (i) F is right continuous,Let X be a continuous random variable with cdf F(x). Show that E(I(X < x)) = F(r) where I is the indicator function (1 if XLet X1, X2,..., Xn be independent exponential variables, parameter λ. Show by induction that X2 S = X₁ + X₂ + + Xn has the r(λ, n) distribution.Let X have pdf = { (a) Determine the distribution function Fx (x), by firstly finding k. fx kr(1-x), 0≤x≤ 1; otherwise.Let Q be a continuous random variable with PDF | 6q(1 – q) if 0 < q < 1 fo(q) = otherwise This Q represents the probability of success of a Bernoulli random variable X, i.e., P(X =1|Q = q) = q. Find foix (q|x) for x E {0, 1} and all q.8. Assume that X is a continuous random variable with pdf 1 if -6SEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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