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
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