Question

prove the chebyshev theorem using the classical definition of variance.

Answer 1

I'm just posting in case someone provides an answer, I want to see too.

Answer 2

looking up the theorem on the net seems to suggest there are several, which one?
Bertrand's postulate
Chebyshev's inequality
Chebyshev's sum inequality
Chebyshev's equioscillation theorem
The statement that if the function has a limit at infinity, then the limit is 1 (where π is the prime-counting function). This result has been superseded by the prime number theorem.

Answer 3

the chebychev's inequality more precisely

Answer 4

no I don't, I just want to see the answer. There are probably only a handful on this site that can do this problem. Here's one now!

Answer 5

It would help a lot of if you give exactly a statement of what it is you're trying to prove and what exactly you want to use as hypothesis.

Answer 6

\[P( \left| X- \mu \right| < k \sigma) \ge 1-1/k ^{2} \]

Answer 7

\[\sigma \neq 0\]

Answer 8

Still not exactly clear. X is an arbitrary random variable, or a Gaussian, or something else?

Answer 9

X is a random variable
k>0

Answer 10

"Discrete Probability" by Hugh Gordon has a proof that appears to use variance.

Answer 11

for this proof I've got only its proof by applying Markov's inequality but I'm required to prove it using the classical definition of variance

Answer 12

i think this may be what you are looking for

Answer 13

@satellite73 thnx but it does not seem to be that what I am look for

Answer 14

Ok, given Markov's inequality, here's a simple proof.

X is a random variable with mean \( \mu \) and variance \(\sigma^2\).

Define a new random variable \(Y = (X - \mu)^2\). By definition of variance, the expectation of Y,
\[ E[Y] = Var(X) = \sigma^2 \]

Now
\[ P( |X - \mu| \geq k\sigma) = P( Y \geq k^2\sigma^2 ), \ \ \hbox{by definition of Y} \]
\[ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \leq \frac{1}{k^2\sigma^2} E[Y], \ \ \ \hbox{by Markov's inequality } . \]
\[ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1/k^2\]

Hence \( P( |X - \mu| < k\sigma) = 1 - 1/k^2 \).

Answer 15

That last equality should be inequality.

We have that
\[ P(|X - \mu| \geq k\sigma) \leq 1/k^2 \]
Hence
\[ P(|X - \mu| < k\sigma) \geq 1 - 1/k^2 \]

Answer 16

So one more bite at the apple. You want a proof that does not use Markov's inequality?

Answer 17

i thought you wanted one that uses variance. the one i sent uses it.