find zero state response y[n+4]-y[n]=x[n], if x[n]= e^-n u[n]

Expert Solution

Manojponduru answered 1 year ago

Let X ∼ Poisson(λ) and Y ∼ U[X, 2X]. Find E(Y ) and V ar(Y ).

define U=x+y, V=x-y. find the joint and marginal pdf of U and V

Let X and Y be continuous random variables with E[X] = E[Y] = 4 and var(X)...

Let X and Y be continuous random variables with E[X] = E[Y] = 4 and var(X) = var(Y) = 10. A new random variable is defined as: W = X+2Y+2. a. Find E[W] and var[W] if X and Y are independent. b. Find E[W] and var[W] if E[XY] = 20. c. If we find that E[XY] = E[X]E[Y], what do we know about the relationship between the random variables X and Y?

Given: E[x] = 4, E[y] = 6, Var(x) = 2, Var(y) = 1, and cov(x,y) =...

Given: E[x] = 4, E[y] = 6, Var(x) = 2, Var(y) = 1, and cov(x,y) = 0.2 Find a lower bound on (5 < x + y < 10). State the theorem used.

hx=u, hy=u/2. Find the Marshallian demands for X and Y.

Consider the ODE y"+ 4 y'+ 4 y = 5 e^(− 2 x ). ( a)...

Consider the ODE y"+ 4 y'+ 4 y = 5 e^(− 2 x ). ( a) Verify that y 1 ( x) = e − 2 x and y 2 ( x) = xe − 2 x satisfy the corresponding homogeneous equation. (b) Use the Superposition Principle, with appropriate coefficients, to state the general solution y h ( x ) of the corresponding homogeneous equation. (c) Verify that y p ( x) = 52 x 2 e − 2 x...

x(n)=(1/4)|n|, find x(ω)

E(Y) = 4, E(X) = E(Y+2), Var(X) = 5, Var(Y) = Var(2X+2) a. What is E(2X...

E(Y) = 4, E(X) = E(Y+2), Var(X) = 5, Var(Y) = Var(2X+2) a. What is E(2X -2Y)? b. What is Var(2X-Y+2)? c. What is SD(3Y-3X)? 2) In a large community, 65% of the residents want to host an autumn community fair. The rest of the residents answer either disagree or no opinion on this proposal. A sample of 18 residents was randomly chosen and asked for their opinions. a) What is th probability that the residents answer either disagree or...

Solve x(y^2+U)Ux -y(x^2+U)Uy =(x^2-y^2)U, U(x,-x)=1

That is, PDEs of the general form A(x, y, u) ∂u(x, y) ∂x + B(x, y,...

That is, PDEs of the general form A(x, y, u) ∂u(x, y) ∂x + B(x, y, u) ∂u(x, y) ∂y = C(x, y, u), (1) for some A, B and C. To solve such PDEs we first find characteristics, curves in the solution space (x, y, u) parametrically given by (x(τ ), y(τ ), u(τ )), which satisfy dx dτ = A(x, y, u), dy dτ = b(x, y, u), du dτ = C(x, y, u). (2) We find solutions...

Question