Deviation (specific number)

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In summary, the conversation discusses ways to select a number T that minimizes the sum of squared errors between T and all the elements in a set of random real numbers. Three possible approaches are suggested, including using the vertex of the set, finding the derivative of the function, or using the sample mean. It is noted that the same approach may not work for minimizing the deviation of all elements in the set with respect to T, and that the mean is the minimizing value for mean squared error. The conversation also explores the effect of incorporating different factors in the calculation of T.
  • #1
onako
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Given a set of random real numbers G=(a, b, c, d, e, f, ...), I am supposed to calculate the number T such that the sum of squared errors between that
number and all the elements in G is mimimum. Intuitively, this should be the average of the numbers in G, but I am not sure how to proceed with the proof (perhaps I am missing something important here).

Would the same answer hold for the question: compute a number T such that the deviation of all the elements in G with respect to T is mimimum. (The standard deviation is with respect to the mean value, but is there some other value T such that the deviation with respecto to it is mimimum (smaller that standard deviation and all others))
In addition, how would the conclusions relate to the median value. Again, intuitively, this would match the number T with the closest value in G.

Your opinion on this is highly appreciated. Thanks
 
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  • #2
I'm assuming your set G is finite with n elements. Your problem is to select T so that

[tex]
S(T) = \sum_{x \in G} (x-T)^2
[/tex]

is minimized as a function of T. You have (at least) three ways to proceed.

a) S(T) is a quadratic function in T.

[tex]
S(T) = \sum_{x \in G} (x^2 - 2xT + T^2) = \sum_{x \in G} x^2 -2T \sum_{x \in G} x + n T^2
[/tex]

By definition the vertex of G gives the minimum value, so use -b/(2a) to find the first coordinate of the vertex

b) S(T) is a differentiable function of T (it's a polynomial);find the derivative, set equal to zero, and solve for T

c) Note that

[tex]
S(T) = \sum_{x \in G} (x-T)^2 = \sum_{x \in G} (x - \overline x + \overline x - T)^2 = \sum_{x \in G} (x-\overline x)^2 + 2\sum_{x \in G} (x-\overline x)(\overline x - T) + \sum_{x \in G} (\overline x - T)^2
[/tex]

You can show the middle term is zero; this means that S(T) is the sum of two non-negative
quantities. How can you pick T to make the the sum of those pieces as small as possible?

"Would the same answer hold for the question: compute a number T such that the deviation of all the elements in G with respect to T is mimimum. (The standard deviation is with respect to the mean value, but is there some other value T such that the deviation with respecto to it is mimimum (smaller that standard deviation and all others)) "
No - as long as you measure mean squared error the mean is the minimizing value.

If you choose to minimize

[tex]
\sum_{x \in G} |x - T|
[/tex]

the solution is T = median of numbers.
 
  • #3
Many thanks for your message.

for c)
minimization of the expression assumes exclusion of the third term, which is set to 0 with T equal to mean value. This means that the mean IS actually the minimizer of the sum of squared errors.

However, I wonder about "compute a number T such that the deviation of all the elements in G with respect to T is mimimum."
 
  • #4
Actually all three of the approaches I mentioned lead to the conclusion that the choice T = sample mean will minimize g(T).

"However, I wonder about "compute a number T such that the deviation of all the elements in G with respect to T is mimimum."

Not sure what you mean here - that's what the work we've discussed does.
 
  • #5
Ok. It means S(T) = sigma^2*N, sigma denoting the standard deviation and N the number of elements in G. What does it say about two sets G1 and G2, having the same standard deviation and same number of elements; they would have same S(T1) and S(T2), although their means might be different?
In other words, given two sets G1 and G2 of equal length, what condition needs be satisfied for S(T1)=S(T2); they would need to have same standard deviation, or same mean?

Thanks
 
Last edited:
  • #6
I tried to incorporate the effect of different factors depending on x entries, say f(x)=x^p

[tex]
S(T) = \sum_{x \in G} f(x)(x-T)^2
[/tex]

The individual effect is, for example, [tex] x_1^p(x_1-T)^2[/tex]. If I'm doing it correctly, not the minimum is the weighted average:

T = (f(x1)x1 + f(x2)x2 + ... + f(xn)xn ) / (f(x1) + f(x2) + ... + f(xn))

Please correct me if I'm wrong.
 
  • #7
I tried to incorporate the effect of different factors depending on x entries, say f(x)=x^p

[tex]
S(T) = \sum_{x \in G} f(x)(x-T)^2
[/tex]

The individual effect is, for example, [tex] x_1^p(x_1-T)^2[/tex]. If I'm doing it correctly, the minimum is the weighted average:

T = (f(x1)x1 + f(x2)x2 + ... + f(xn)xn ) / (f(x1) + f(x2) + ... + f(xn))

Please correct me if I'm wrong.
 

What is deviation?

Deviation is a measure of how much a set of data varies from its mean or average value. It tells us how spread out the data points are from the average.

How is deviation calculated?

Deviation is calculated by taking the difference between each data point and the mean, squaring those differences, finding the average of those squared differences, and then taking the square root of that average. The formula for deviation is: √[(x1 - x̄)^2 + (x2 - x̄)^2 + ... + (xn - x̄)^2] / n

What is the difference between standard deviation and variance?

Standard deviation and variance are both measures of how spread out a set of data is, but they differ in the unit of measurement. Standard deviation is measured in the same units as the data, while variance is measured in squared units.

Why is deviation important in statistics?

Deviation is important in statistics because it helps us understand the distribution of data and how much it varies from the average. It is also used in many statistical calculations, such as calculating confidence intervals and performing hypothesis tests.

What does a high or low deviation indicate about the data?

A high deviation indicates that the data points are spread out from the mean, while a low deviation indicates that the data points are clustered closely around the mean. A higher deviation can also suggest that the data is more diverse or has more outliers, while a lower deviation can suggest that the data is more consistent and has fewer outliers.

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