Cantelli's Inequality and Chebyshev's Inequality

In summary, the two equations found using Chebyshev's inequality give an upper bound of 0.5625 for the number of customers that would be in the store on a given day if it had a mean of 80 and a variance of 225.
  • #1
whitejac
169
0

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?
 
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  • #2
whitejac said:

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).
 
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  • #3
Samy_A said:
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.
 
  • #4
whitejac said:

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.
 
  • #5
Ray Vickson said:
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:
whitejac said:

Homework Equations


P(|x - Ex| ≥ b) ≤ σ2/b2
P(X ≥ a) ≤ σ2 / (σ2 + a2)
 
Last edited:
  • #6
whitejac said:
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.
whitejac said:
"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).
 

1. What is Cantelli's Inequality?

Cantelli's Inequality, also known as the Chebyshev-Cantelli Inequality, is a statistical theorem that provides bounds on the probability that a random variable deviates from its mean by a certain amount.

2. How is Cantelli's Inequality related to Chebyshev's Inequality?

Cantelli's Inequality is a special case of Chebyshev's Inequality, which is a more general result in probability theory. Cantelli's Inequality assumes that the random variable has a finite variance, whereas Chebyshev's Inequality applies to any random variable with a finite mean and variance.

3. What is the significance of Cantelli's Inequality in statistics?

Cantelli's Inequality is an important tool in probability theory and statistics, as it provides a way to quantify the likelihood of deviations from the mean in a random variable. This allows for the calculation of confidence intervals and the evaluation of the accuracy of statistical models.

4. How is Cantelli's Inequality used in real-world applications?

Cantelli's Inequality is commonly used in fields such as finance, economics, and engineering to analyze data and make predictions. It is also used in quality control and risk management to assess the deviation of a process from its target value.

5. Are there any limitations to Cantelli's Inequality?

Like any statistical theorem, Cantelli's Inequality has its limitations. It assumes that the random variable is independent and identically distributed, which may not always be the case in real-world data. Additionally, the bounds provided by Cantelli's Inequality may be too wide in some cases, leading to less precise estimates of the probability of deviations.

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