I don't understand why this formula gives the mean of a pdf. I don't really understand pdf's too well, but I do understand that the probability of exactly x realization occurring, is zero (at least conceptually, but perhaps not mathematically). I understand partly (I guess?) that the probability of x occurring within a given interval can be defined. But then how is it, that multiplying by x on this function, suddenly gives the mean (after doing the integration)? I've tried figuring it in my head but can't manage it

0le
 Follow
 2 followers
 20 badges
 Send a private message to 0le
 Thread Starter
Offline20ReputationRep: Follow
 1
 23102015 17:33
Last edited by 0le; 23102015 at 17:35. 
 Follow
 2
 23102015 17:50
(Original post by djpailo)
I don't understand why this formula gives the mean of a pdf. I don't really understand pdf's too well, but I do understand that the probability of exactly x realization occurring, is zero (at least conceptually, but perhaps not mathematically). I understand partly (I guess?) that the probability of x occurring within a given interval can be defined. But then how is it, that multiplying by x on this function, suddenly gives the mean (after doing the integration)? I've tried figuring it in my head but can't manage it
that is why the formula is identical to that of finding the x coordinate of the centre of mass of a uniform lamina 
0le
 Follow
 2 followers
 20 badges
 Send a private message to 0le
 Thread Starter
Offline20ReputationRep: Follow
 3
 23102015 18:07
(Original post by TeeEm)
the mean is the weighted average, i.e the balancing point, i.e the "centre of mass" of a lamina whose shape is that of the PDF.
that is why the formula is identical to that of finding the x coordinate of the centre of mass of a uniform lamina
Is it something like this example:
http://www.mathwords.com/w/weighted_average.htm
But done on infinitesimal little bits or intervals? So is it like saying we have a score in our homework of around 92, a grade of around 68 etc and for each of those "arounds" or better worded, [i]intervals[/]i, we are assigned a weight which, when integrated across, gives the probability? 
 Follow
 4
 23102015 18:12
(Original post by djpailo)
Still a bit confused ;(
Is it something like this example:
http://www.mathwords.com/w/weighted_average.htm
But done on infinitesimal little bits or intervals? So is it like saying we have a score in our homework of around 92, a grade of around 68 etc and for each of those "arounds" or better worded, [i]intervals[/]i, we are assigned a weight which, when integrated across, gives the probability?
I wish I could be more helpful. 
 Follow
 5
 23102015 18:23
(Original post by djpailo)
I don't understand why this formula gives the mean of a pdf. I don't really understand pdf's too well, but I do understand that the probability of exactly x realization occurring, is zero (at least conceptually, but perhaps not mathematically). I understand partly (I guess?) that the probability of x occurring within a given interval can be defined. But then how is it, that multiplying by x on this function, suddenly gives the mean (after doing the integration)? I've tried figuring it in my head but can't manage it
If you read the first page, it will explain it a little bit.
I was asking myself this question 5months ago, but I forgot 
MathsAndChess
 Follow
 4 followers
 11 badges
 Send a private message to MathsAndChess
Offline11ReputationRep: Follow
 6
 23102015 18:26
(Original post by djpailo)
I don't understand why this formula gives the mean of a pdf. I don't really understand pdf's too well, but I do understand that the probability of exactly x realization occurring, is zero (at least conceptually, but perhaps not mathematically). I understand partly (I guess?) that the probability of x occurring within a given interval can be defined. But then how is it, that multiplying by x on this function, suddenly gives the mean (after doing the integration)? I've tried figuring it in my head but can't manage it
Assume a discrete function Where is the probability of x occurring, The part of the sum that will contribute to the Expected value must be , as this is the mean outcome for this variables value. Therefore it is logical to assume that
Finite Integration shows the area under a curve. But the area is really just lots of little rectangles. Which you should have explored through the Trapezium Rule. Therefore we can assume integration is like "continuous addition." This therefore implies we can apply the same the same logic to our continuous function:
Assuming
Last edited by MathsAndChess; 23102015 at 21:35. 
 Follow
 7
 23102015 18:39
Am a 3rd year LSE statistician so will try to explain.
Say you had a distribution, which could only take 2 values. 0 or 1 each with equal probability 0.5. To calculate the mean you would do 0.5*0+0.5*1, which is the summation of xf(x). Say now you can have values 0, 0.5 and 1. You would do a third times 0 plus a third times 0.5 plus a third times 1. Now if you have a continuous distribution you can have infinite values so you have an infinitely small number times every number between 0 and 1 added together, which is the integral of X times the probability of X happening f(x) between all values hence the infinity and minus infinity. So yeah weighted average done on an infinitely small scale.
(Original post by djpailo)
Still a bit confused ;(
Is it something like this example:
http://www.mathwords.com/w/weighted_average.htm
But done on infinitesimal little bits or intervals? So is it like saying we have a score in our homework of around 92, a grade of around 68 etc and for each of those "arounds" or better worded, [i]intervals[/]i, we are assigned a weight which, when integrated across, gives the probability? 
0le
 Follow
 2 followers
 20 badges
 Send a private message to 0le
 Thread Starter
Offline20ReputationRep: Follow
 8
 26102015 10:18
(Original post by LMSZ)
Am a 3rd year LSE statistician so will try to explain.
Say you had a distribution, which could only take 2 values. 0 or 1 each with equal probability 0.5. To calculate the mean you would do 0.5*0+0.5*1, which is the summation of xf(x). Say now you can have values 0, 0.5 and 1. You would do a third times 0 plus a third times 0.5 plus a third times 1. Now if you have a continuous distribution you can have infinite values so you have an infinitely small number times every number between 0 and 1 added together, which is the integral of X times the probability of X happening f(x) between all values hence the infinity and minus infinity. So yeah weighted average done on an infinitely small scale. 
 Follow
 9
 26102015 10:28
(Original post by djpailo)
Thanks, this helps. Do you have a similar explanation for variance which would be x^2p(x) instead? 
0le
 Follow
 2 followers
 20 badges
 Send a private message to 0le
 Thread Starter
Offline20ReputationRep: Follow
 10
 26102015 13:25
(Original post by LMSZ)
(x^2)p(x) isn't the variance that is the expected value of x^2. E(x^2)E(x)^2 is the variance.
http://www.turbulenceonline.com/Pub...anuary2013.pdf
I thought it was a "central moment" if you find it about a mean, and just a "moment" if you use the formula on page 25. 
 Follow
 11
 26102015 13:30
(Original post by djpailo)
I see where you are coming from, but I'm confused as to why, on pdf page 25, the formula for the nth moment is this:
http://www.turbulenceonline.com/Pub...anuary2013.pdf
I thought it was a "central moment" if you find it about a mean, and just a "moment" if you use the formula on page 25. 
0le
 Follow
 2 followers
 20 badges
 Send a private message to 0le
 Thread Starter
Offline20ReputationRep: Follow
 12
 26102015 13:40
(Original post by LMSZ)
I'm confused what you mean. The formula on the article you sent me finds the moment as in E(X), E(X^2). The central moment is E(XE(X))^n so the second central moment is the variance. 
DFranklin
 Follow
 60 followers
 17 badges
 Send a private message to DFranklin
Offline17ReputationRep: Follow
 13
 26102015 16:43
(Original post by djpailo)
eq 2.17 has c^n * function and says that those are higher order moments, hence variance, skewness etc :s 
0le
 Follow
 2 followers
 20 badges
 Send a private message to 0le
 Thread Starter
Offline20ReputationRep: Follow
 14
 26102015 17:53
(Original post by DFranklin)
Looking at the document, it seems to clearly distinguish between moments and central moments. There's a brief moment where it talks about the variance without stating it's a central moment, but it doesn't state that it's a moment either, so I don't really see grounds for confusion.
Say you had a distribution, which could only take 2 values. 0 or 1 each with equal probability 0.5. To calculate the mean you would do 0.5*0+0.5*1, which is the summation of xf(x). Say now you can have values 0, 0.5 and 1. You would do a third times 0 plus a third times 0.5 plus a third times 1. Now if you have a continuous distribution you can have infinite values so you have an infinitely small number times every number between 0 and 1 added together, which is the integral of X times the probability of X happening f(x) between all values hence the infinity and minus infinity. So yeah weighted average done on an infinitely small scale. 
DFranklin
 Follow
 60 followers
 17 badges
 Send a private message to DFranklin
Offline17ReputationRep: Follow
 15
 26102015 19:06
(Original post by djpailo)
So I'm wondering what the how to picture c^n*f(x) instead of c*f(x) where the latter was explained like this:
As far as relating c^n to the explanation: it's exactly the same, only instead of finding the average value of x f(x), you're finding the average value of x^n f(x).
I do think it's a lot easier to understand these things in the finite case (where you are summing discrete probabilities) before progressing to the infinite one (where you need to integrate). 
0le
 Follow
 2 followers
 20 badges
 Send a private message to 0le
 Thread Starter
Offline20ReputationRep: Follow
 16
 26102015 21:11
(Original post by DFranklin)
First off, those equations aren't correct. You are visualising c^n f(c) (or x^n f(x)). You should not be mixing c and x here.
As far as relating c^n to the explanation: it's exactly the same, only instead of finding the average value of x f(x), you're finding the average value of x^n f(x).
I do think it's a lot easier to understand these things in the finite case (where you are summing discrete probabilities) before progressing to the infinite one (where you need to integrate).
0*0.5 + 1*0.5 = 0.5 mean.
0*0*0.5 + 1*1*0.5 = 0.5 variance
or a dice
1*(1/6) + 2*(1/6) + 3*(1/6) +4*(1/6) + 5*(1/6) + 6*(1/6) = 3.5 = mean
1*(1/6)*(1/6) + 2*(1/6)*(1/6) + 3*(1/6)*(1/6) +4*(1/6)*(1/6) + 5*(1/6)*(1/6) + 6*(1/6)*(1/6) = 0.58 variance
are these correct? I just find it strange that the formulae work like they do. I'm so use to just finding the arithmetic mean, so when finding the mean like this, it feels unnatural.
From the pdf, can I then interpret b_x(c) as the probability that the random variable, x, falls between a class width of c if we had a discrete case?
Then making the jump to infinite case, is that the probability that the random variable, x, falls between an infinitesimally small class width? 
DFranklin
 Follow
 60 followers
 17 badges
 Send a private message to DFranklin
Offline17ReputationRep: Follow
 17
 26102015 21:43
(Original post by djpailo)
Okay, so in discrete case going to the coin example stated earlier:
0*0.5 + 1*0.5 = 0.5 mean.
0*0*0.5 + 1*1*0.5 = 0.5 variance
or a dice
1*(1/6) + 2*(1/6) + 3*(1/6) +4*(1/6) + 5*(1/6) + 6*(1/6) = 3.5 = mean
1*(1/6)*(1/6) + 2*(1/6)*(1/6) + 3*(1/6)*(1/6) +4*(1/6)*(1/6) + 5*(1/6)*(1/6) + 6*(1/6)*(1/6) = 0.58 variance
You have found when the correct formula for E[X^2] is .
I don't want to seem patronizing, but you really need to find an elementary book covering this  it is S1/S2 material. Most undergrad level books will be assuming you already know this to some extent and so will be skimming over it at an unrealistic pace. 
0le
 Follow
 2 followers
 20 badges
 Send a private message to 0le
 Thread Starter
Offline20ReputationRep: Follow
 18
 26102015 23:01
(Original post by DFranklin)
No, what you have calculated here is E[X^2], which is not the same as the variance. The variance is E((XE(X))^2), which can be shown (via some manipulation that I suspect you are not ready for) to also equal E[X^2]  E[X]^2.
In this case you haven't even managed to calculate E[X^2] correctly.
You have found when the correct formula for E[X^2] is .
I don't want to seem patronizing, but you really need to find an elementary book covering this  it is S1/S2 material. Most undergrad level books will be assuming you already know this to some extent and so will be skimming over it at an unrealistic pace.
You've said E[X^2] is not the variance. Okay, if that is the case, what is the difference between the moment and central moment of variance, if E[X^2] is not the moment (or better worded, why is that not the variance if the mean is zero)?
What good books would you suggest?
EDIT:
I recalculated the variance, is it 2.92 for the dice?Last edited by 0le; 26102015 at 23:20. 
 Follow
 19
 28102015 12:54
(Original post by TeeEm)
the mean is the weighted average, i.e the balancing point, i.e the "centre of mass" of a lamina whose shape is that of the PDF.
that is why the formula is identical to that of finding the x coordinate of the centre of mass of a uniform lamina
 Stats PDF probability density function
 Difference between cumulative distribution function & ...
 Probability density function help
 Probability Density Function
 Probability HELP!!!
 Intergration In Probability Density Functions
 Statistics help: Probability density functions
 S2 Standardizing the Normal Distribution
 Help with this S2 question really struggling :(
 Finding the median using the PDF

University of Bath

Economics, Statistics and Mathematics
Queen Mary University of London

Mathematics and Financial Economics
University of Dundee

Mathematics and Computer Science
University of Liverpool

University of Kent

University of Oxford

Mathematics and Physics (with Year Abroad)
University of Bath

University of Oxford

University of Oxford

Mathematics and Theoretical Physics
University of St Andrews
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
 SherlockHolmes
 Notnek
 charco
 Mr M
 Changing Skies
 F1's Finest
 rayquaza17
 RDKGames
 davros
 Gingerbread101
 Kvothe the Arcane
 TeeEff
 The Empire Odyssey
 Protostar
 TheConfusedMedic
 nisha.sri
 claireestelle
 Doonesbury
 furryface12
 Amefish
 harryleavey
 Lemur14
 brainzistheword
 Rexar
 Sonechka
 TheAnxiousSloth
 EstelOfTheEyrie
 CoffeeAndPolitics
 an_atheist
 Labrador99
 EmilySarah00