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
Q1. A group of five applicants for a pair of identical jobs consists of three men and two
women. The employer is to select two of the five applicants for the jobs. Let S
denote the set of all possible outcomes for the employer's selection. Let A denote
the subset of outcomes corresponding to the selection of two men and B the subset
corresponding to the selection of at least one woman. List the outcomes in A, B,
AUB, AN B, and An B. (Denote the different men and women by M₁, M2, M3
and W₁, W2, respectively.)
Q3 (8 points)
Q3. A survey classified a large number of adults according to whether they were diag-
nosed as needing eyeglasses to correct their reading vision and whether they use
eyeglasses when reading. The proportions falling into the four resulting categories
are given in the following table:
Use Eyeglasses for Reading
Needs glasses Yes
No
Yes
0.44
0.14
No
0.02
0.40
If a single adult is selected from the large group, find the probabilities of the events
defined below. The adult
(a) needs glasses.
(b) needs glasses but does not use them.
(c) uses glasses whether the glasses are needed or not.
4. (i) Let a discrete sample space be given by
N = {W1, W2, W3, W4},
and let a probability measure P on be given by
P(w1) = 0.2, P(w2) = 0.2, P(w3) = 0.5, P(wa) = 0.1.
Consider the random variables X1, X2 → R defined by
X₁(w1) = 1, X₁(w2) = 2,
X2(w1) = 2, X2 (w2) = 2,
Find the joint distribution of X1, X2.
(ii)
X1(W3) = 1, X₁(w4) = 1,
X2(W3) = 1, X2(w4) = 2.
[4 Marks]
Let Y, Z be random variables on a probability space (, F, P).
Let the random vector (Y, Z) take on values in the set [0, 1] x [0,2] and let the
joint distribution of Y, Z on [0, 1] x [0,2] be given by
1
dPy,z (y, z) ==(y²z+yz2) dy dz.
harks 12 Find the distribution Py of the random variable Y.
[8 Marks]
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