Help with an expectation value formula

[Agrippa]

Imagine a particle in an equally weighted superposition of being located in three distant regions P, Q, and R, and imagine you stand in region P with a measuring device. The probability of finding the particle there is 1/3.

Now imagine a large number N of particles prepared in that same state. I'm told that in region P we should expect to find N_P = (N±√N)/3 particles.

My question: Does anyone know the name of this formula so that I can google it and study it? I'm puzzled as to why we shouldn't just expect N/3 particles.

 

[Orodruin]

There is a difference between expectation value and "how many events you expect". The expectation value is N/3, but if N is very large it is rather unlikely to get exactly N/3. The number after the ± tells you the standard deviation (although in this case it is not correct, the standard deviation of the binomial distribution with p=1/3 is sqrt(2N)/3, not sqrt(N)/3) and therefore the typical interval you would expect to fall in.

 

[blue_leaf77]

if N is very large

Do you mean when N is relatively small ($N\pm \sqrt{N} \approx N$ when $N$ is very large) ?

 

[Orodruin]

Do you mean when N is relatively small ($N\pm \sqrt{N} \approx N$ when $N$ is very large) ?

No. I mean exactly what I wrote. The probability of getting exactly N/3 is (1/3)^[N/3](2/3)^[2N/3] N!/[(N/3)!(2N/3)!] if N is divisible by 3 and zero otherwise. This goes to zero as N increases.

Edit: In fact, in absolute terms, the standard deviation grows with N. It is only the relative error that decreases.

 

[Agrippa]

There is a difference between expectation value and "how many events you expect". The expectation value is N/3, but if N is very large it is rather unlikely to get exactly N/3. The number after the ± tells you the standard deviation (although in this case it is not correct, the standard deviation of the binomial distribution with p=1/3 is sqrt(2N)/3, not sqrt(N)/3) and therefore the typical interval you would expect to fall in.

Thanks that's very helpful! Do you know if this equation has a name? I would like to be able to look it up and learn more about it. For example, I was puzzled as to why sqrt(N) is the standard deviation. But I'm also puzzled as to why it should be sqrt(2N) e.g. where does the 2 come from?

 

[Orodruin]

Thanks that's very helpful! Do you know if this equation has a name? I would like to be able to look it up and learn more about it. For example, I was puzzled as to why sqrt(N) is the standard deviation. But I'm also puzzled as to why it should be sqrt(2N) e.g. where does the 2 come from?

From computing the standard deviation of the binomial distribution. Bin(N,p) has expectation value pN and variance p(1-p)N. These are properties of the distribution and will differ between distributions.

 

[Agrippa]

From computing the standard deviation of the binomial distribution. Bin(N,p) has expectation value pN and variance p(1-p)N. These are properties of the distribution and will differ between distributions.

Okay thanks that's helpful, I've been looking through links about binomial distribution and the formulas for variance and standard deviation.
The link states that the binomial distribution has the following properties:

• The mean of the distribution (μx) is equal to n * P .
• The http://stattrek.com/Help/Glossary.aspx?Target=Variance (σ²x) is n * P * ( 1 - P ).
• The http://stattrek.com/Help/Glossary.aspx?Target=Standard%20deviation (σx) is sqrt[ n * P * ( 1 - P ) ].

In our case P = 1/3. So variance is n*1/3*(1 - 1/3) = n*0.2222.. and standard deviation is sqrt[n*1/3*(1 - 1/3)] = sqrt[n*0.2222..].

But then the question still remains: where does the 2 come from in your calculations? How do you get 2 from 0.2222..?

 

[Heinera]

Orodruin just simplified the formula for the case P=1/3. Plug in the same N in both your and Orodruin's expressions. Do you find a difference?

 

[Agrippa]

Orodruin just simplified the formula for the case P=1/3. Plug in the same N in both your and Orodruin's expressions. Do you find a difference?

The difference between my expression:
N_P = (N±√N)/3
And Orodruin's expression:
N_P = (N±√[2N])/3
Is that N is multiplied by 2 inside the sqrt.
So for same N of course there will be different outputs.
The question is why either formula is "correct".
Orodruin's argument for the correctness of Orodruin's formula involves "computing the standard deviation of the binomial distribution". My previous post expressed why I do not understand that argument (i.e. where the 2 comes from).

 

[Heinera]

I meant you should compare Orodruin's expression to your latest, i.e. sqrt[n*1/3*(1 - 1/3)] = sqrt[n*0.2222..].

 

[Agrippa]

I meant you should compare Orodruin's expression to your latest, i.e. sqrt[n*1/3*(1 - 1/3)] = sqrt[n*0.2222..].

Okay, so let N = 2.

Then (my latest):
sqrt[N*0.2222..] = 0.666666..

And (Orodruin's expression):
sqrt(2N))/3 = 0.666666..

Meanwhile (my post-1 expression):
(sqrt(N))/3 = 0.47.

So the idea is that we should use:
N_P = (N±√[2N])/3
...because for N particles, each with 1/3 prob of collapsing in P, we should (in P) expect to see N/3 particles plus or minus the standard deviation?

Well if so, that certainly helps. Thanks! Now I need study why standard deviation gives the formula that is significant for our expectations.

 

[Orodruin]

I think you are trying to assign a meaning that simply is not there. The mean and standard deviations are two numbers that tell you something about a distribution, namely what the average would be after a large number of trials and the typical deviation from that result. It is pure convention to write it A±B and you cannot find any deeper meaning than "A is the mean and B the standard deviation". There is nothing to derive here.

 

[Orodruin]

Okay, so let N = 2.

Then (my latest):
sqrt[N*0.2222..] = 0.666666..

And (Orodruin's expression):
sqrt(2N))/3 = 0.666666..

Meanwhile (my post-1 expression):
(sqrt(N))/3 = 0.47.

There is no need to look at a particular N nor write out the ratio in decimal form. (1/3)(1-1/3)=2/9. Take the square root of that and you have $\sqrt 2/3$.