Expected value of random variable^4 greater than variance^2
up vote
3
down vote
favorite
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 8 hours ago
J.G.
20.6k21933
asked 8 hours ago
user603569
597
add a comment |
up vote
3
down vote
favorite
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 8 hours ago
J.G.
20.6k21933
asked 8 hours ago
user603569
597
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
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 8 hours ago
J.G.
20.6k21933
asked 8 hours ago
user603569
597
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 8 hours ago
J.G.
20.6k21933
asked 8 hours ago
user603569
597
edited 8 hours ago
J.G.
20.6k21933
asked 8 hours ago
user603569
597
edited 8 hours ago
J.G.
20.6k21933
edited 8 hours ago
J.G.
20.6k21933
edited 8 hours ago
J.G.
20.6k21933
20.6k21933
asked 8 hours ago
user603569
597
asked 8 hours ago
user603569
597
asked 8 hours ago
user603569
597
597
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
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 8 hours ago
drhab
95.6k543126
add a comment |
up vote
2
down vote
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 8 hours ago
J.G.
20.6k21933
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
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 8 hours ago
drhab
95.6k543126
add a comment |
up vote
3
down vote
accepted
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 8 hours ago
drhab
95.6k543126
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
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 8 hours ago
drhab
95.6k543126
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 8 hours ago
drhab
95.6k543126
answered 8 hours ago
drhab
95.6k543126
answered 8 hours ago
drhab
95.6k543126
answered 8 hours ago
drhab
95.6k543126
95.6k543126
add a comment |
add a comment |
up vote
2
down vote
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 8 hours ago
J.G.
20.6k21933
add a comment |
up vote
2
down vote
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 8 hours ago
J.G.
20.6k21933
add a comment |
up vote
2
down vote
up vote
2
down vote
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 8 hours ago
J.G.
20.6k21933
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 8 hours ago
J.G.
20.6k21933
answered 8 hours ago
J.G.
20.6k21933
answered 8 hours ago
J.G.
20.6k21933
answered 8 hours ago
J.G.
20.6k21933
20.6k21933
add a comment |
add a comment |
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