Prove the Cauchy-Schwarz inequality, namely, ( E [ X Y ] ) 2 ≤ E [ X 2 ] E [ Y 2 ] Hint: Unless Y = − t X for some constant, in which case the inequality holds with equality, it follows that for all 0 < E [ ( t X + Y ) 2 ] = E [ X 2 ] t 2 + 2 E [ X Y ] t + E [ Y 2 ] Hence, the roots of the quadratic equation E [ X 2 ] t 2 + 2 E [ X Y ] t + E [ Y 2 ] = 0 must be imaginary, which implies that the discriminant of this quadratic equation must be negative.
Prove the Cauchy-Schwarz inequality, namely, ( E [ X Y ] ) 2 ≤ E [ X 2 ] E [ Y 2 ] Hint: Unless Y = − t X for some constant, in which case the inequality holds with equality, it follows that for all 0 < E [ ( t X + Y ) 2 ] = E [ X 2 ] t 2 + 2 E [ X Y ] t + E [ Y 2 ] Hence, the roots of the quadratic equation E [ X 2 ] t 2 + 2 E [ X Y ] t + E [ Y 2 ] = 0 must be imaginary, which implies that the discriminant of this quadratic equation must be negative.
Solution Summary: The author calculates the Thecauchy schwarz inequality using a sequence of independent and identically distributed random vectors.
Prove the Cauchy-Schwarz inequality, namely,
(
E
[
X
Y
]
)
2
≤
E
[
X
2
]
E
[
Y
2
]
Hint: Unless
Y
=
−
t
X
for some constant, in which case the inequality holds with equality, it follows that for all
0
<
E
[
(
t
X
+
Y
)
2
]
=
E
[
X
2
]
t
2
+
2
E
[
X
Y
]
t
+
E
[
Y
2
]
Hence, the roots of the quadratic equation
E
[
X
2
]
t
2
+
2
E
[
X
Y
]
t
+
E
[
Y
2
]
=
0
must be imaginary, which implies that the discriminant of this quadratic equation must be negative.
Formula Formula A polynomial with degree 2 is called a quadratic polynomial. A quadratic equation can be simplified to the standard form: ax² + bx + c = 0 Where, a ≠ 0. A, b, c are coefficients. c is also called "constant". 'x' is the unknown quantity
QUESTION 18 - 1 POINT
Jessie is playing a dice game and bets $9 on her first roll. If a 10, 7, or 4 is rolled, she wins $9. This happens with a probability of . If an 8 or 2 is rolled, she loses her $9. This has a probability of J. If any other number is rolled, she does not win or lose, and the game continues. Find the expected value for Jessie on her first roll.
Round to the nearest cent if necessary. Do not round until the final calculation.
Provide your answer below:
5 of 5
(i) Let a discrete sample space be given by
Ω = {ω1, 2, 3, 4},
Total marks 12
and let a probability measure P on be given by
P(w1) 0.2, P(w2) = 0.2, P(w3) = 0.5, P(w4) = 0.1.
=
Consider the random variables X1, X2 → R defined by
X₁(w3) = 1, X₁(4) = 1,
X₁(w₁) = 1, X₁(w2) = 2,
X2(w1) = 2, X2(w2) = 2, X2(W3) = 1, X2(w4) = 2.
Find the joint distribution of X1, X2.
(ii)
[4 Marks]
Let Y, Z be random variables on a probability space (N, F, P).
Let the random vector (Y, Z) take on values in the set [0,1] × [0,2] and let the
joint distribution of Y, Z on [0,1] × [0,2] be given by
1
dPy,z(y, z)
(y²z + y²²) dy dz.
Find the distribution Py of the random variable Y.
[8 Marks]
Total marks 16
5.
Let (,,P) be a probability space and let X : → R be a random
variable whose probability density function is given by f(x) = }}|x|e¯|×| for
x Є R.
(i)
(ii)
Find the characteristic function of the random variable X.
[8 Marks]
Using the result of (i), calculate the first two moments of the
random variable X, i.e., E(X") for n = 1, 2.
(iii) What is the variance of X?
[6 Marks]
[2 Marks]
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