Stats - Geometric Variance Proof

In summary, to prove the geometric variance equation using algebra only, you can use the concept of Zeno's Paradox to rewrite the equation in a form that uses infinite sums. You can then use the concept of Zeno's Paradox to show that the infinite sum of terms on the right side of the equation is equal to the finite sum of terms on the left side of the equation, thus proving the equation to be true.
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Stats -- Geometric Variance Proof

Hi,
I'm a student in South-East Indiana, enrolled in a AP Stats class.
Our teacher has asked us to prove the geometric variance equation (the first equation pictured) USING ALGEBRA ONLY.
I've gotten it all the way down to the 2nd equation and now I'm stuck, I've called MANY homework help lines and none can help.
It's due March 12th, next Wednesday, so I have about a week.
Please, your my last resort, give me a hint!

He hinted it has something to do with infinity, and something called Zeno's Paradox, and I've googled Zeno's Paradox but all I've been able to find is about motion not existing.

Note: No calculus is to be used, this class is a pre-req for calculus.

Thanks for the help.
 

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The proof of the geometric variance equation (the first equation pictured) USING ALGEBRA ONLY can be done by using the concept of Zeno's Paradox. Zeno's paradox states that motion cannot exist, and this can be used to prove the geometric variance equation. To do this, you need to rewrite the geometric variance equation in a form that uses infinite sums. This can be done by rewriting the equation as: Var(X) = p^2 * (1 + 2p + 3p^2 + 4p^3 + ... + np^n) - [p + 2p^2 + 3p^3 + 4p^4 + ... + np^(n+1)]^2Next, you need to use the concept of Zeno's Paradox to show that this equation equals the original geometric variance equation. Using the concept of Zeno's Paradox, you can show that the infinite sum of terms on the right side of the equation is equal to the finite sum of terms on the left side of the equation. This can be done by showing that the terms on the right side can be broken up into an infinite number of smaller and smaller terms, each of which is equal to the finite sum of terms on the left side of the equation. For example, the second term on the right side of the equation can be rewritten as: 2p^2 = p^2 + p^2 By breaking up the second term into two smaller terms, each of which is equal to the first term on the left side of the equation, we can prove that the infinite sum of terms on the right side of the equation is equal to the finite sum of terms on the left side of the equation. By continuing to break up the terms on the right side of the equation into smaller and smaller terms, each of which is equal to the corresponding terms on the left side of the equation, you can prove that the equation is indeed true.
 
  • #3


Hello,

I can certainly understand your struggle with this proof, as it can be challenging to prove without using calculus. However, I am glad to see that you have reached out for help and are determined to find a solution.

Firstly, let's recap the geometric variance equation:

Var(X) = E[(X - μ)^2] = ∑(x - μ)^2 * P(X = x)

To prove this, we must use the definition of variance, which is the average of the squared differences between each data point and the mean. So, we need to show that the formula above is equivalent to:

Var(X) = E[X^2] - μ^2

Now, let's think about what the right side of the equation represents. E[X^2] is the expected value of X squared, and μ^2 is the square of the mean. This is similar to the formula for calculating variance in a finite population, where we take the sum of squared differences between each data point and the mean.

However, in this case, we are dealing with an infinite population, which is where the hint about infinity and Zeno's Paradox comes into play. Zeno's Paradox states that if you continuously divide a distance in half, you will never reach the end. In this case, we can think of our infinite population as the continuous division of data points.

So, when we calculate E[X^2], we are essentially taking the sum of all possible squared data points in an infinite population. Similarly, μ^2 represents the mean of all possible squared data points. Therefore, the difference between these two values gives us the average of the squared differences, which is the definition of variance.

I hope this helps to give you some direction in your proof. Remember to carefully consider the properties of infinite populations and how they relate to the formula for variance. Best of luck on your assignment!
 

1. What is a geometric variance proof?

A geometric variance proof is a mathematical method used to demonstrate the relationship between the variance of a set of values and their geometric mean.

2. How is the geometric variance proof calculated?

The geometric variance proof is calculated by taking the logarithm of each value in the set, finding the average of these values, and then taking the exponential of this average. This result is then subtracted from the squared value of the geometric mean of the original set.

3. What is the significance of the geometric variance proof?

The geometric variance proof is important because it helps to quantify the spread or variability of a set of values around their geometric mean. This can be useful in analyzing data and making statistical inferences.

4. How does the geometric variance proof differ from the traditional variance proof?

The traditional variance proof is calculated by taking the squared difference between each value and the arithmetic mean of the set. The geometric variance proof, on the other hand, takes the logarithm of each value and uses the geometric mean. This makes it more suitable for data sets with large variations.

5. Can the geometric variance proof be used for non-normal distributions?

Yes, the geometric variance proof can be used for non-normal distributions as it is not affected by extreme values or outliers. However, it is important to note that it may not be appropriate for all types of data and other methods may be more suitable in some cases.

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