Question

Given a random variable X has the following pmf:

X

-1

0

1

P[X]

0.25

0.5

0.25

Define Y = X2 & W= Y+2.

Which one of the following statements is not true?

A) V[Y] = 0.25.

B) E[XY] = 0.

C) E[X3] = 0.

D) E[X+2] = 2.

E) E[Y+2] = 2.5.

F) E[W+2] = 4.5.

G) V[X+2] = 0.5.

H) V[W+2] = 0.25.

I) P[W=1] = 0.5

J) X and W are not independent.

Answer #1

**please rate me high.**

(a) TRUE / FALSE If X is a random variable, then (E[X])^2 ≤
E[X^2]. (b) TRUE / FALSE If Cov(X,Y) = 0, then X and Y are
independent. (c) TRUE / FALSE If P(A) = 0.5 and P(B) = 0.5, then
P(AB) = 0.25. (d) TRUE / FALSE There exist events A,B with P(A)not
equal to 0 and P(B)not equal to 0 for which A and B are both
independent and mutually exclusive. (e) TRUE / FALSE Var(X+Y) =
Var(X)...

Let the random variable X and Y have the joint pmf f(x, y) =
xy^2/c where x = 1, 2, 3; y = 1, 2, x + y ≤ 4 , that is, (x, y) are
{(1, 1),(1, 2),(2, 1),(2, 2),(3, 1)} .
(a) Find c > 0 .
(b) Find μX
(c) Find μY
(d) Find σ^2 X
(e) Find σ^2 Y
(f) Find Cov (X, Y )
(g) Find ρ , Corr (X, Y )
(h) Are X...

Let X be a random variable with the pmf p(x) which is
positive at x=1;0;1, and zero elsewhere. If E(X^3) = 0 andE(X^2)
=p(0),what is p(1)?

Consider two random variable X and Y with joint PMF given in the
table below.
Y = 2
Y = 4
Y = 5
X = 1
k/3
k/6
k/6
X = 2
2k/3
k/3
k/2
X = 3
k
k/2
k/3
a) Find the value of k so that this is a valid PMF. Show your
work.
b) Re-write the table with the joint probabilities using the
value of k that you found in (a).
c) Find the marginal...

(a) It is given that a random variable X such that P(X =−1) =P(X
= 1) = 1/4, P(X = 0) = 1/2. Find the mgf of X, mX(t)
(b) Let X1 and X2 be two iid random varibles such that P(Xi
=1)=P(Xi =−1)=1/2, i=1,2. Use the mgfs to prove that X and Y
=(X1+X2)/2 have the same distribution

Let X be a discrete random variable with the pmf
p(x): 0.8 for x=-4,
0.1 for x=-2,
0.07 for x=0,
0.03 for x=2
a) Find E(2/X)
b) Find E(lXl)
c) Find Var(lXl)

A. (i) Consider the random variable X with pmf: pX (−1) = pX (1)
= 1/8, pX (0) = 3/4.
Show that the Chebyshev inequality P (|X − μ| ≥ 2σ) ≤ 1/4 is
actually an equation for
this random variable.
(ii) Find the pmf of a different random variable Y that also takes
the values {−1, 0, 1}
for which the Chebyshev inequality P (|X − μ| ≥ 3σ) ≤ 1/9 is
actually an equation.

Let X be a discrete random variable with probability mass
function (pmf) P (X = k) = C *ln(k) for k = e; e^2 ; e^3 ; e^4 ,
and C > 0 is a constant.
(a) Find C.
(b) Find E(ln X).
(c) Find Var(ln X).

Suppose a random variable X takes on the value of -1 or 1, each
with the probability of 1/2. Let y=X1+X2+X3+X4, where X1,....X4 are
independent. Find E(Y) and Find Var(Y)

1. Suppose a random variable X has a probability density
function
f(x)= {cx^2 -1<x<1,
{0 otherwise
where c > 0.
(a) Determine c.
(b) Find the cdf F ().
(c) Compute P (-0.5 < X < 0.75).
(d) Compute P (|X| > 0.25).
(e) Compute P (X > 0.75 | X > 0).
(f) Compute P (|X| > 0.75| |X| > 0.5).

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 12 minutes ago

asked 16 minutes ago

asked 20 minutes ago

asked 28 minutes ago

asked 57 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago

asked 2 hours ago