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What is Var(X) for my defined X?

by Pzi
Tags: defined, varx
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Pzi
#1
Jan11-13, 01:20 PM
P: 26
This applies to natural numbers [itex]n[/itex] and [itex]N[/itex] where [itex]n<N[/itex].

We have [itex]N[/itex] balls representing numbers [itex]1,2,...,N[/itex].
We randomly choose [itex]n[/itex] of those balls which happen to represent numbers [itex]{k_1},{k_2},...,{k_n}[/itex].
We then define a random variable [itex]X = {k_1} + {k_2} + ... + {k_n}[/itex].
What is the mean and variation of [itex]X[/itex]?

Well there are [itex]\left( {\begin{array}{*{20}{c}}N\\n\end{array}} \right)[/itex] equally likely combinations and every one of them brings [itex]n[/itex] summands. So we get [itex]\left( {\begin{array}{*{20}{c}}
N\\
n
\end{array}} \right) \cdot n[/itex] summands overall. Since there is no bias towards any particular number it means that every number is added [itex]\left( {\begin{array}{*{20}{c}}
N\\
n
\end{array}} \right) \cdot \frac{n}{N}[/itex] times hence our mean:
[itex]E\left( X \right) = \frac{{\left( {\begin{array}{*{20}{c}}
N\\
n
\end{array}} \right) \cdot \frac{n}{N} \cdot \left( {1 + 2 + ... + N} \right)}}{{\left( {\begin{array}{*{20}{c}}
N\\
n
\end{array}} \right)}} = \frac{n}{N} \cdot \frac{{N\left( {N + 1} \right)}}{2} = \frac{{n\left( {N + 1} \right)}}{2}[/itex].

Unfortunately variance seems to be an entirely different animal since I cannot just rip sums apart and add numbers in a different order. Any ideas?
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mathman
#2
Jan11-13, 03:28 PM
Sci Advisor
P: 6,104
It is a straightforward, but messy, calculation.

Let X = ∑ ki. mean = E(X), var = E(X2) - (E(X))2
To compute E(X) all you need is E(ki) = N/2.

To compute E(X2), some work is required.
You need (a) E(kikj) for i ≠ j, and (b) E(ki2).
For (a) I got N(N+1)/4 - (2N+1)/6, for (b) I got (N+1)(2N+1)/6.

To get the final answers you need to multiply the mean by n.
For the second moment {E(X2)}, there are n(n-1) (a) terms and n (b) terms.

Good luck!
mathman
#3
Jan14-13, 04:18 PM
Sci Advisor
P: 6,104
In case you weren't able to work it out, I got the following:

E(X) = n(N+1)/2
Var(X) = n(N+1)(N-n)/12

See if you get same.

Pzi
#4
Jan17-13, 09:16 AM
P: 26
What is Var(X) for my defined X?

Quote Quote by mathman View Post
In case you weren't able to work it out, I got the following:

E(X) = n(N+1)/2
Var(X) = n(N+1)(N-n)/12

See if you get same.
I appreciate your dedication very much. Sorry for not getting back to this topic in a reasonable amount of time.

So for i≠j you apparently got E(kikj) = N(N+1)/4 - (2N+1)/6.
Meanwhile I got
[tex]E({k_i}{k_j}) = \frac{1}{{\left( {\begin{array}{*{20}{c}}
N\\
2
\end{array}} \right)}}\sum\limits_{i = 1}^{N - 1} {i \cdot \sum\limits_{j = i + 1}^N j } = \frac{2}{{N(N - 1)}}\sum\limits_{i = 1}^{N - 1} {i \cdot \frac{{(N + i + 1)(N - i)}}{2}} = \frac{{N(N + 1)}}{2} + \frac{{1 - 2N}}{6} + \frac{{N(1 - N)}}{4}[/tex]

Is my logic flawed here? Also I checked the literature for answers and can confirm that your conclusion about mean and variation is correct.
mathman
#5
Jan17-13, 03:57 PM
Sci Advisor
P: 6,104
Quote Quote by Pzi View Post
I appreciate your dedication very much. Sorry for not getting back to this topic in a reasonable amount of time.

So for i≠j you apparently got E(kikj) = N(N+1)/4 - (2N+1)/6.
Meanwhile I got
[tex]E({k_i}{k_j}) = \frac{1}{{\left( {\begin{array}{*{20}{c}}
N\\
2
\end{array}} \right)}}\sum\limits_{i = 1}^{N - 1} {i \cdot \sum\limits_{j = i + 1}^N j } = \frac{2}{{N(N - 1)}}\sum\limits_{i = 1}^{N - 1} {i \cdot \frac{{(N + i + 1)(N - i)}}{2}} = \frac{{N(N + 1)}}{2} + \frac{{1 - 2N}}{6} + \frac{{N(1 - N)}}{4}[/tex]

Is my logic flawed here? Also I checked the literature for answers and can confirm that your conclusion about mean and variation is correct.
You made at least one arithmetic error. The summand in the second summation should be
i(N-i+1)(N-i)/2. I'll let you finish from there.

In any case I got E(X2) = n(n-1)(N+1)(3N+2)/12 + n(N+1)(2N+1)/6,
where the first term is the contribution of all products of different numbers and the second term is the contribution of all the squares.
Pzi
#6
Jan17-13, 04:58 PM
P: 26
Quote Quote by mathman View Post
You made at least one arithmetic error. The summand in the second summation should be
i(N-i+1)(N-i)/2. I'll let you finish from there.
Why do you say so?
(i+1)+(i+2)+(i+3)...+(N-2)+(N-1)+N = (N+i+1)(N-i)/2

Quote Quote by mathman View Post
In any case I got E(X2) = n(n-1)(N+1)(3N+2)/12 + n(N+1)(2N+1)/6,
where the first term is the contribution of all products of different numbers and the second term is the contribution of all the squares.
Well the first term is E(kikj) added n(n-1) times isn't it? Hence E(kikj)=(N+1)(3N+2)/12 (for i≠j) which is not the same as previously stated by yourself E(kikj)=N(N+1)/4-(2N+1)/6.
Sorry if I missed something.

P.S. http://www.wolframalpha.com/input/?i...i%3D1%2CN-1%29
it gives the same result as mine so I'm confused now, is it the wrong way to calculate E(kikj) ?
mathman
#7
Jan18-13, 07:10 PM
Sci Advisor
P: 6,104
Sorry - I sometimes get sloppy in my arithmetic (Jan. 11 - my error was in dividing by N(N+1) when I should have
divided by (N-1)N), including my comment saying you had an error. However the calculation on Jan. 14 was done very carefully (where I divided by N(N-1)). This included the result I mentioned yesterday.

I can't figure out why yours comes out differently.
My calculation - E(XiXj) = {(∑i)2 - ∑i2}/N(N-1), where the sums are i = (1,N).
mathman
#8
Jan18-13, 07:35 PM
Sci Advisor
P: 6,104
One quick check, for N = 2, the product of different x's ≡ 2, so the expectation is obviously 2.

In any case, your result is the same as mine.
Pzi
#9
Jan20-13, 03:22 PM
P: 26
Thanks for all the help.

To sum things up I'd just like to say that I almost got it right after your first post. Unfortunately I made a very silly mistake at the very end. Which in turn led to a whole phase of contemplation and doubts.

I attach that almost-correct handwritten solution just for fun.
Differences in notation: E(X)=M(X), Var(X)=D(X), X=W.
Attached Thumbnails
failure.jpg  


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