# Thread: Chebyshev's Inequality problem

1. ## Chebyshev's Inequality problem

Let X ~ Poisson (lambda). With Chebyshev's Inequality, show that

P(X >=2 lambda) <= 1/lambda

The way I thought to interpret this theorem was to determine what percentage of the data is clustered near the mean when faced with a non-normal distribution like a Poisson distribution that determines the probability of a given number of events within a fixed interval. This is where I get tripped up when attempting to lay out the proof with a symbol (lambda) as opposed to a number. Any assistance would be greatly appreciated. Thanks.

2. ## Re: Chebyshev's Inequality problem

Note that the mean and standard deviation of the Poisson random variable are both lambda. So, look back at Chebyshev and see where these values fit in.

P(|x-mu|<k*(sigma)) >= 1-(1/k^2)

If we subtract this probability from 1 we should get P([-k*(sigma) >= x-mu] or [x-mu >= k*(sigma)]) ...

Excuse my excessive use of brackets.

3. ## Re: Chebyshev's Inequality problem

Pardon me, the variance is lambda

4. ## Re: Chebyshev's Inequality problem

I thought the formula stated for r.v. X with E(X) = mu and alpha > 0 ,

P[|x - mu| >= alpha] <= Var(X)/alpha^2

This is where I get confused about plugging in the Poisson distribution characteristics.

5. ## Re: Chebyshev's Inequality problem

I think you are very close. There are a few ways to see the same inequality https://en.wikipedia.org/wiki/Chebyshev%27s_inequality

Look under the section "Probabilistic statement"
Perhaps yours is correct. I haven't fiddled around with it too much

Our goal is to get the absolute value bars out of the problem. So, we need to restate/rearrange this inequality to deal with that. If you want, you can treat the lambda as a number. Then switch back later.

6. ## Re: Chebyshev's Inequality problem

Let alpha = lambda in your version. almost qed

7. ## Re: Chebyshev's Inequality problem

Ok, P(|X - mu| >=k*sigma) <= 1/k^2

So if the variance for Poission is lambda, let's plug in.

P(|X-mu| >=2(lambda)k) <= 1/k^2

P(|X-mu| <= (lambda)k) >= 1- 1/k^2

P(-k(lambda) >= X - mu

Let me know if I did the proof right.

Thanks.

8. ## Re: Chebyshev's Inequality problem

Originally Posted by greg6363
P[|x - mu| >= alpha] <= Var(X)/alpha^2
I was thinking:

P.S. I hope is your favorite Greek letter.

9. ## Re: Chebyshev's Inequality problem

Here goes.

P(|X - lambda| >= lambda) <= lambda/lambda^2

P(|X - lambda| >= 2 lambda) <= 1/lambda

P(|X - lambda| <= lambda >= 1 - 1/lambda

P(-lambda >= X - lambda)

Let me know. Thanks.

10. ## Re: Chebyshev's Inequality problem

is equivalent to or

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