Suppose the random variables X and Y have a pdf given by f (x, y) = kæy on 0
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A: Hello! As you have posted 2 different questions, we are answering the first question. In case you…
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A: The probability density function of Y is given below: fYy|θ=2θ2θ-y, for y∈0,θ
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A: Given information:
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A: The PDF is a probability that a random variable, say X, will take a value exactly equal to x.
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A: The PDF of (X1, X2) is fX1,X2(x1,x2)=12πe-12(x12+x22),(x1,x2)∈R2Let X1=Y1sin Y2 and X2=Y1cos Y2Then,…
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- Let random variables X and Y have the joint pdf fX,Y (x, y) = 4xy, 0 < x < 1, 0 < y < 1 0, otherwise Find the joint pdf of U = X^2 and V = XY.Find E(R) and V (R) for a random variable R whose moment-generating function ismR(t) = e2t(1-3t2)-1The joint PDF of the random variables X and Y is constant on the shaded region, as shown in the image below, and is zero outside. It can be determined that fX,Y(x,y)=2/3. Determine E[XY]. (The answer is not 3/4)
- Your internal body temperature T in °F is a Gaussian (μ =98.6, σ = 0.4) random variable. In terms of the Φ function, find P[T > 100]. Does this model seem reasonable?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-Y2Let X1 ... Xn i.i.d random variables with Xi ~ U(0,1). Find the pdf of Q = X1, X2, ... ,Xn. Note that first that -log(Xi) follows exponential distribuition.
- Q4) If X is a continuous random variable having pdf ke~ (2x+3y) x>0y>0 xy) = = e p(x) { 0 otherwise Find a) the constant k b) P(X>1) ¢) X, X2, 02, standard deviation.Let a random variable X has the CDF as;Fx(x) = 1 ; x ≥ 1 1/2+x ; 0 ≤ x < 1 0 ; x < 0 Verify that Fx(x) is a CDF. Also- (i) find the PDF of X (ii) Calculate E[ex ] (iii) Determine the probability P[X = 0|X ≤ 0.5]Find the maximum likelihood estimator for θ in the pdf f(y; θ) = 2y/(1 − θ^2), θ ≤ y ≤ 1.
- Consider a random variable Y with PDF Pr(Y=k)=pq^(k-1),k=1,2,3,4,5....compute for E(2Y)let X and Y be a random variables having pdf f(x,y)=2xy 0<x<y<1 Find P(X/Y<1/2)If we let RX(t) = ln MX(t), show that R X(0) = μ and RX(0) = σ2. Also, use these results to find the mean and the variance of a random variable X having the moment-generating function MX(t) = e4(et−1)