The fourth moment of a centered random variable is at least equal to the square of its variance
Let X be a random variable with mean µ and variance $sigma^2$. Show that
$E(X-mu)^4geq sigma^4$
and use this to show that the kurtosis of $X$ is at least $-2$.
This looks like a form of Chebyshev's equation:
$P(|X-mu|geq a) leq frac{sigma^2}{a^2}$
But does not relate the inequality in the probability.
I tried expanding $E(X-mu)^4$ out, but that seemed like a dead end.
Can someone guide me on how to solve this? If it does deal with Chebyshev, can someone explain in detail how to get the inequality out of the absolute value probability?
random-variables expected-value
edited Dec 12 at 17:06
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Let X be a random variable with mean µ and variance $sigma^2$. Show that
$E(X-mu)^4geq sigma^4$
and use this to show that the kurtosis of $X$ is at least $-2$.
This looks like a form of Chebyshev's equation:
$P(|X-mu|geq a) leq frac{sigma^2}{a^2}$
But does not relate the inequality in the probability.
I tried expanding $E(X-mu)^4$ out, but that seemed like a dead end.
Can someone guide me on how to solve this? If it does deal with Chebyshev, can someone explain in detail how to get the inequality out of the absolute value probability?
random-variables expected-value
edited Dec 12 at 17:06
Did
246k23220454
asked Dec 9 at 16:14
user603569
728
add a comment |
Let X be a random variable with mean µ and variance $sigma^2$. Show that
$E(X-mu)^4geq sigma^4$
and use this to show that the kurtosis of $X$ is at least $-2$.
This looks like a form of Chebyshev's equation:
$P(|X-mu|geq a) leq frac{sigma^2}{a^2}$
But does not relate the inequality in the probability.
I tried expanding $E(X-mu)^4$ out, but that seemed like a dead end.
Can someone guide me on how to solve this? If it does deal with Chebyshev, can someone explain in detail how to get the inequality out of the absolute value probability?
random-variables expected-value
edited Dec 12 at 17:06
Did
246k23220454
asked Dec 9 at 16:14
user603569
728
Let X be a random variable with mean µ and variance $sigma^2$. Show that
$E(X-mu)^4geq sigma^4$
and use this to show that the kurtosis of $X$ is at least $-2$.
This looks like a form of Chebyshev's equation:
$P(|X-mu|geq a) leq frac{sigma^2}{a^2}$
But does not relate the inequality in the probability.
I tried expanding $E(X-mu)^4$ out, but that seemed like a dead end.
Can someone guide me on how to solve this? If it does deal with Chebyshev, can someone explain in detail how to get the inequality out of the absolute value probability?
random-variables expected-value
random-variables expected-value
edited Dec 12 at 17:06
Did
246k23220454
asked Dec 9 at 16:14
user603569
728
edited Dec 12 at 17:06
Did
246k23220454
asked Dec 9 at 16:14
user603569
728
edited Dec 12 at 17:06
Did
246k23220454
edited Dec 12 at 17:06
Did
246k23220454
edited Dec 12 at 17:06
Did
246k23220454
246k23220454
asked Dec 9 at 16:14
user603569
728
asked Dec 9 at 16:14
user603569
728
asked Dec 9 at 16:14
user603569
728
728
add a comment |
add a comment |
2 Answers
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Setting $Y=X-mu$ it must actually be shown that $mathbb EY^4geq(mathbb EY^2)^2$.
This is a direct consequence of: $$mathbb EY^4-(mathbb EY^2)^2=mathsf{Var}Y^2geq0$$
answered Dec 9 at 16:20
drhab
96.5k544127
add a comment |
You don't need Chebyshev. Note that $E(X-mu)^4=operatorname{Var}(X-mu)^2+E^2(X-mu)^2$. The first term $ge 0$; the second is $sigma^4$.
answered Dec 9 at 16:19
J.G.
21.7k21934
add a comment |
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2 Answers
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2 Answers
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active
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active
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active
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votes
Setting $Y=X-mu$ it must actually be shown that $mathbb EY^4geq(mathbb EY^2)^2$.
This is a direct consequence of: $$mathbb EY^4-(mathbb EY^2)^2=mathsf{Var}Y^2geq0$$
answered Dec 9 at 16:20
drhab
96.5k544127
add a comment |
Setting $Y=X-mu$ it must actually be shown that $mathbb EY^4geq(mathbb EY^2)^2$.
This is a direct consequence of: $$mathbb EY^4-(mathbb EY^2)^2=mathsf{Var}Y^2geq0$$
answered Dec 9 at 16:20
drhab
96.5k544127
add a comment |
Setting $Y=X-mu$ it must actually be shown that $mathbb EY^4geq(mathbb EY^2)^2$.
This is a direct consequence of: $$mathbb EY^4-(mathbb EY^2)^2=mathsf{Var}Y^2geq0$$
answered Dec 9 at 16:20
drhab
96.5k544127
Setting $Y=X-mu$ it must actually be shown that $mathbb EY^4geq(mathbb EY^2)^2$.
This is a direct consequence of: $$mathbb EY^4-(mathbb EY^2)^2=mathsf{Var}Y^2geq0$$
answered Dec 9 at 16:20
drhab
96.5k544127
answered Dec 9 at 16:20
drhab
96.5k544127
answered Dec 9 at 16:20
drhab
96.5k544127
answered Dec 9 at 16:20
drhab
96.5k544127
96.5k544127
add a comment |
add a comment |
You don't need Chebyshev. Note that $E(X-mu)^4=operatorname{Var}(X-mu)^2+E^2(X-mu)^2$. The first term $ge 0$; the second is $sigma^4$.
answered Dec 9 at 16:19
J.G.
21.7k21934
add a comment |
You don't need Chebyshev. Note that $E(X-mu)^4=operatorname{Var}(X-mu)^2+E^2(X-mu)^2$. The first term $ge 0$; the second is $sigma^4$.
answered Dec 9 at 16:19
J.G.
21.7k21934
add a comment |
You don't need Chebyshev. Note that $E(X-mu)^4=operatorname{Var}(X-mu)^2+E^2(X-mu)^2$. The first term $ge 0$; the second is $sigma^4$.
answered Dec 9 at 16:19
J.G.
21.7k21934
You don't need Chebyshev. Note that $E(X-mu)^4=operatorname{Var}(X-mu)^2+E^2(X-mu)^2$. The first term $ge 0$; the second is $sigma^4$.
answered Dec 9 at 16:19
J.G.
21.7k21934
answered Dec 9 at 16:19
J.G.
21.7k21934
answered Dec 9 at 16:19
J.G.
21.7k21934
answered Dec 9 at 16:19
J.G.
21.7k21934
21.7k21934
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add a comment |
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