The bottleneck here is to proof that the sum of N gaussian iid random variables each one divided by N follows also a normal distribution, which by the way, it's assumed as black magic here. It would be necessary thefore to perferm a N convolution of each random variables this way: Y(x1, x2...xN)=(1/N^N)*CONV(x1, x2...xN)
I’m not sure if that was assumed at any point… I think they only talk about the expectation and variance of the sample mean, but don’t assert that it too is a normal distribution (perhaps the central limit thm helps with that?)
I think that is taken care of by Central Limit Theorem, which states even if we r sampling from a population with unknown distribution, the sampling distribution of Mn will still be approximately normal with mean as meu and variance square(Sigma)/n, provided sample size (n) is large enough.
barney boy In Australia, Saturday night lotto went from 40 ball in 1972 then from 1985 the numbers were increased to 45 balls. Too many winners as the number of entries had increased. In this case there is to me a force in the tumbler that decides what numbers will be drawn it is mind blowing (to me) Let me explain when the 40 balls were drawn they worked to the law of large numbers, this we would expect. With the remaining 5 balls that were added 13 years later you would expect a lower count of them coming up if we plot them over 33 years (This period is when the extra 5 balls were added). This is how they fall in September 2018, from 1972 till today the lowest number drawn is 17 (seventeen) 285 times and the highest number drawn is 11 (eleven) 345 times. For the 5 balls 41 to 45 that were added in 1985 the lowest drawn number is 44 , at 283 and the highest is 42 at 319 times. The 5 balls that were added later are at times exceeding the original 40 balls. The 5 balls added 13 years later were drawn approximately 672 less than the original 40 balls. To me this says these 5 balls added later have been drawn out at an extremely higher rate than the original 40 balls. but the average total now for the 45 balls is very even. To a person who did not know that 5 balls were inserted 672 draws later would say this shows the laws of large numbers is correct. The above information says this is not true. Try explaining that. To me the analogy is we have 2 cars, the first car drives for 20 km, the second car is waiting at the 20 km mark. They then both drive together to the 100 km point. The first car has driven 100 km the second car has driven 80 km but the odometers on each car show 100 km. I cannot see how this can be worked out mathematically.
The way I think of it is the noise doesn't have any rhyme or reason and is just as likely to inflate the true value as it is to reduce it, so the more times you repeat the experiment the more likely this noise is to cancel itself out. If you flip a coin 4 times you could get 4 heads, but if you flip it 10 billion times you will not have 10 billion heads, it will almost certainly be very close to 5 billion, because there's been sufficient chance for the noise to cancel itself out.
Yes that is also correct. The (n) in the denominator is replaced by (n-1) as a sort of correction. It's called Bessel's correction. It's alright to skip that for the purpose of proof as done here.
Wait, the bessel's correction and the (n-1) term is seen in Sample Variance S² S² = [1/(n-1)] Σ (Xᵢ - Sample mean)². In the video, it is the sample mean that's being discussed, and it has the usual denominator n.
I don't think there can ever be a better explanation than these. Detailed explanation.
thank you so much. I've watched all the videos so far and I have a much better understanding now.
THE GOAT!
great explanation with clear voice
Oooo boy that greek accent made my day. Ευχαριστω for the perfect explanation.
The bottleneck here is to proof that the sum of N gaussian iid random variables each one divided by N follows also a normal distribution, which by the way, it's assumed as black magic here. It would be necessary thefore to perferm a N convolution of each random variables this way: Y(x1, x2...xN)=(1/N^N)*CONV(x1, x2...xN)
I’m not sure if that was assumed at any point… I think they only talk about the expectation and variance of the sample mean, but don’t assert that it too is a normal distribution (perhaps the central limit thm helps with that?)
I think that is taken care of by Central Limit Theorem, which states even if we r sampling from a population with unknown distribution, the sampling distribution of Mn will still be approximately normal with mean as meu and variance square(Sigma)/n, provided sample size (n) is large enough.
barney boy
In Australia, Saturday night lotto went from 40 ball in 1972 then from 1985 the numbers were increased to 45 balls. Too many winners as the number of entries had increased. In this case there is to me a force in the tumbler that decides what numbers will be drawn it is mind blowing (to me) Let me explain when the 40 balls were drawn they worked to the law of large numbers, this we would expect. With the remaining 5 balls that were added 13 years later you would expect a lower count of them coming up if we plot them over 33 years (This period is when the extra 5 balls were added). This is how they fall in September 2018, from 1972 till today the lowest number drawn is 17 (seventeen) 285 times and the highest number drawn is 11 (eleven) 345 times. For the 5 balls 41 to 45 that were added in 1985 the lowest drawn number is 44 , at 283 and the highest is 42 at 319 times. The 5 balls that were added later are at times exceeding the original 40 balls. The 5 balls added 13 years later were drawn approximately 672 less than the original 40 balls. To me this says these 5 balls added later have been drawn out at an extremely higher rate than the original 40 balls. but the average total now for the 45 balls is very even. To a person who did not know that 5 balls were inserted 672 draws later would say this shows the laws of large numbers is correct. The above information says this is not true. Try explaining that. To me the analogy is we have 2 cars, the first car drives for 20 km, the second car is waiting at the 20 km mark. They then both drive together to the 100 km point. The first car has driven 100 km the second car has driven 80 km but the odometers on each car show 100 km. I cannot see how this can be worked out mathematically.
Oh
Nicely explained.
It's very much the best
E[ Xi/n ] = p P( |Mn - p| >= e ) -> 0 as n gets large enough. But, know the caveat.
Керемет♥
Why does the noisy element have 0 as the expected value?
E[Xi]=mu so we can considere that Xi=mu +Wi and scince E[Xi]=mu +E[Wi] then E[Wi]=0
The way I think of it is the noise doesn't have any rhyme or reason and is just as likely to inflate the true value as it is to reduce it, so the more times you repeat the experiment the more likely this noise is to cancel itself out. If you flip a coin 4 times you could get 4 heads, but if you flip it 10 billion times you will not have 10 billion heads, it will almost certainly be very close to 5 billion, because there's been sufficient chance for the noise to cancel itself out.
6:00
Holy…
Should sample mean be the sum of observations divided by (n-1)?
Yes that is also correct. The (n) in the denominator is replaced by (n-1) as a sort of correction. It's called Bessel's correction. It's alright to skip that for the purpose of proof as done here.
Wait, the bessel's correction and the (n-1) term is seen in Sample Variance S²
S² = [1/(n-1)] Σ (Xᵢ - Sample mean)².
In the video, it is the sample mean that's being discussed, and it has the usual denominator n.
@@deepanshhh so, the bessel's correction is for the sample Variance, not sample mean.