Cantelli's Inequality and Chebyshev's Inequality

[whitejac]

Homework Statement

The number of customers visiting a store during a day is a random variable with mean EX=100and variance Var(X)=225.

1. Using Chebyshev's inequality, find an upper bound for having more than 120 or less than 80customers in a day. That is, find an upper bound on
   P(X≤80 or X≥120).
2. Using the one-sided Chebyshev inequality (Problem 21), find an upper bound for having

2. Homework Equations
P(|x - Ex| ≥ b) ≤ σ²/b²
P(X ≥ a) ≤ σ² / (σ² + a²)

The Attempt at a Solution

For part 1:
P(X≤80 or X≥120) = P(80 ≥ X ≥120) = P(-20 ≥ X - 100 ≥20) = P(|X-100| ≥ 20)
By Chebychev's:
P(|X-100| ≥ 20) ≥ σ²/b²
thus
P(|X-100| ≥ 20) ≥ 225 / 20² = 0.5625

For part 2:
Using equation 2...
P(X≥121) ≤ 225 / (225 + 121²) = 0.0151

This seems wrong to me because it would seem more reasonable to get a number closer to half of part 1. This answer in part 2 implies that taking just one side of Chebyshev's inequality (the upper half of part 1) and then cutting off the value for 120 reduces it by a half and then almost that entire difference over again... not to mention it imlpies that the majority of the variance isn't even probable. Did I make an arithmetic error?

 

[Samy_A]

Homework Equations

P(|x - Ex| ≥ b) ≤ σ²/b²
P(X ≥ a) ≤ σ² / (σ² + a²)

I think your second relevant equation (the one-sided one) is wrong for random variables with mean<>0.

In your solution for part 1, you have ≥ instead of ≤ (probably a typo).

 

[whitejac]

I think your second relevant equation (the one-sided one) is wrong for random variables with mean<>0.

Oh yes, I see that condition in my book. It's true for Ex = 0 and Var(x) = σ² and any a >0.
So, I think this means that I need to shift my given information to where the mean is 0. I have two ideas about solving this:

Standardize in a similar way to finding the normal distrubution,
We need to begin by standardizing the system to a mean of 0. This is done by (Y - EX) / σ². This should make the mean 0 and standardize the variance to 1? This yields: (121 - 100) / (225) = 0.093.
This was my first instinct, but it doesn't seem like it's going in the correct direction to find "a." Rather it seems like this is headed towards finding it using the normal distribution.

"shift" the mean to 0 by searching for P(X > 120 - 100) = P(X > 20)
which would give me P(X≥21) ≥ (225 / (225 + 21²) = 0.33. This is a much closer answer, however it doesn't shift the variance any.

 

[Ray Vickson]

Homework Statement

The number of customers visiting a store during a day is a random variable with mean EX=100and variance Var(X)=225.

1. Using Chebyshev's inequality, find an upper bound for having more than 120 or less than 80customers in a day. That is, find an upper bound on
   P(X≤80 or X≥120).
2. Using the one-sided Chebyshev inequality (Problem 21), find an upper bound for having

2. Homework Equations
P(|x - Ex| ≥ b) ≤ σ²/b²
P(X ≥ a) ≤ σ² / (σ² + a²)

The Attempt at a Solution

For part 1:
P(X≤80 or X≥120) = P(80 ≥ X ≥120) = P(-20 ≥ X - 100 ≥20) = P(|X-100| ≥ 20)
By Chebychev's:
P(|X-100| ≥ 20) ≥ σ²/b²
thus
P(|X-100| ≥ 20) ≥ 225 / 20² = 0.5625

For part 2:
Using equation 2...
P(X≥121) ≤ 225 / (225 + 121²) = 0.0151

This seems wrong to me because it would seem more reasonable to get a number closer to half of part 1. This answer in part 2 implies that taking just one side of Chebyshev's inequality (the upper half of part 1) and then cutting off the value for 120 reduces it by a half and then almost that entire difference over again... not to mention it imlpies that the majority of the variance isn't even probable. Did I make an arithmetic error?

The statement "P(80 ≥ X ≥120) = something" is nonsense: you cannot have anything bounded below by 120 and bounded above by 80. Anyway, the events {X ≤ 80} and {X ≥ 120} are disjoint, so what does that tell you about P(X ≤ 80 or X ≥ 120)? also: you have the inequalities backwards in Chebychev.

 

[whitejac]

so what does that tell you about P(X ≤ 80 or X ≥ 120)? also: you have the inequalities backwards in Chebychev.

It tells me that I could take 1 - P(80 ≤ X ≤ 120) to find the remaining areas?
Also, the mistake in my inequalities must have been a misclick. I put the correct intentions in the Formulas section:

Homework Equations

P(|x - Ex| ≥ b) ≤ σ2/b2
P(X ≥ a) ≤ σ2 / (σ2 + a2)

 

[Samy_A]

Oh yes, I see that condition in my book. It's true for Ex = 0 and Var(x) = σ² and any a >0.
So, I think this means that I need to shift my given information to where the mean is 0. I have two ideas about solving this:

Standardize in a similar way to finding the normal distrubution,
We need to begin by standardizing the system to a mean of 0. This is done by (Y - EX) / σ².

To standardize the random variable, you have to divide by the standard deviation, not the variance.

"shift" the mean to 0 by searching for P(X > 120 - 100) = P(X > 20)
which would give me P(X≥21) ≥ (225 / (225 + 21²) = 0.33. This is a much closer answer, however it doesn't shift the variance any.

Your notation is sloppy here, as X seems to refer to 2 different random variables, but I see what you meant to write. Anyway, I got the same result by using Cantelli's inequality (see the title of this thread).