Deriving the Mean and Variance of the Sample Mean

[@tuankietnguyenthanh7091]

it's great, I've been kinda confused about this for few days and now finally understand it. Your effort is much appreciated!

[@thekittulegend]

You are a really good educator, Sir you helped me understand very clearly thank you 🙏

[@cutestbear3327]

methodical, clear and concise, supplemented by clean, understandable graphics too. he indeed is a very good educator

[@suleymansahal]

Thank you for these great videos. I understand the logic now but in the past something always confused me, maybe others too. I think the relation between Xbar r.v. and X1,X2,...,Xn r.v.s is not clear. Once you form this relation, the rest is relatively easy. At first I always thought we should incorporate the number of trials in these equations. Later, I figured that out by myself, hopefully correctly. What i understood: Xbar is a new variable which has its own probability distribution and we do not know it. X1, X2 ... Xn (with capital Xs) also have their distributions and their distributions are identical. In each trial we get one observation for Xbar r.v. which is (x1,x2,...,xn)/n. The number of trials makes the distribution more normal. Maybe that point needs a little deeper explanation.

[@jbstatistics]

Thanks for the compliment, and thanks for the feedback. I'm not sure what you think is unclear about the relationship between X bar and X1, X2, ..., Xn, as in the video I define X bar to be (X1+X2+...+Xn)/n. n is the number of observations (I state in the video that X bar is the mean of these n independent observations, then give the formula). As far as the distribution of X bar becoming more normal with increasing sample size, that's a completely different topic than that of this video, and I discuss that notion in great detail in other videos. Cheers.

[@suzyolds2620]

I'm confused about the definition of the Xi's. If the Xi's are independently drawn observations, how can each Xi have a variance? Isn't each Xi a single observation?

[@jbstatistics]

X_1, X2_, ..., X_n are random variables that take on values once we draw the sample. The distribution of X_1 is simply the distribution of the population from which we are drawing the sample. Similarly for X_2 through X_n (provided that they can be thought of as independent random variables, or in other words, that we are sampling independently from the distribution). When I speak of the mean and variance of X_i, I'm not talking about the actual values that we observe once we draw the sample, I'm talking about the theoretical mean and variance of the random variable. This will simply be the mean and variance of the population (distribution) from which we are sampling.

[@sketch1625]

@@jbstatistics It is still hard to understand in simple terms thinking the mean of X_i = Mu and then plugging in this value to the E(X_1)+E(X_2)...E(X_n). Doesn't the Mean/Expected(X_i) = Mu when thinking of ALL random variables (X_i), not each individually? The subbing is not clear to me.

[@yaman4863]

jbstatistics But isn’t E(constant)=constant? Aren’t those X_is random *constants* drawn from the population? Which makes their expectation equal to themselves?

[@yaman4863]

It MAY mean that the Expected value, of any randomly picked variable out of a population, is EXPECTED TO BE the MEAN of that population. Although it’s an obvious remark, that’s the only explanation I can come up with right now.

[@yaman4863]

Turns out that X_1, for instance, is a mean of sample 1. So the Xi itself is not a sample individual, rather, it is the mean of a chunk of individuals(observations). This is more clear in this video: “The Sampling Distribution of the Sample Mean.”

[@GoodLuckForever-wi9kb]

Thanks for explaining this concept in such a simple way.

[@syedahmedali7417]

no doubt you have such a great quality of teaching

[@jbstatistics]

Thanks!

[@sazik]

Can you please help me understand why at 2:39 you say "When a random variable gets multiplied by a constant, its variance gets multiplied by the square of that constant". Can you show a quick proof of that?

[@sparrshnagdda1190]

This is because when a random variable gets multiplied by a constant , the standard deviation will also get affected by that factor. And since, variance is sd squared, it gets squared as well sd= x + y sd= a(x+y).. now that the factor is introduced var=a squared (x+y) squared Hope this helped.

[@MrKorkak]

these are the best videos about statistics and probability... please exponential and gamma dist. to continuos dist.

[@jbstatistics]

Thanks for the compliment! I keep trying to find the time to make videos on the exponential and gamma distributions (as well as many other topics). I should be able to find some time within the next year :)

[@kevinmorfol7912]

omg seeing the proofs makes me feel like i finally get it! thats crazy!

[@jbstatistics]

I'm glad to hear it!

[@aiavicii4243]

are V(X) and V(X bar) the same thing??????

[@kagi95]

Wonderfully clear and professional explanation, thanks!

[@jbstatistics]

kagi95 You are very welcome!

[@harshalshinde227]

+jbstatistics what the heck is the expectation , Iam not getting it, please explain.

[@harshalshinde227]

+jbstatistics dsmpling without replacement mean?

[@nishtharai4521]

Praise God. Thank you for helping me out

[@jeffreylin2488]

@jbstatistics Can the samples of the sampling distribution overlap?

[@gatsu3605]

What would be the variance of the square of the sample mean?

[@LyndaLiu]

I’m confused about one thing… Suppose the sample data are 1,2,3,4,5, then the expected value of X bar = E[(1+2+3+4+5)/5]=1/5*E(1+2+3+4+5)=1/5*[E(1)+E(2)+…+E(5)], how is E(1) the same as E(2)? How can we write E(Xi)=mu? In this case, mu=3, but E(1)=1? So how can we say the sum of the individual expected values are the same and are all equal to mu and therefore n*mu? Thank you!

[@jbstatistics]

If you have sample data, then you have sample data and that does not tell you what the expected value of anything is. You don't know mu. You don't know the expectation anything. The sample data might help us to *estimate* the expected value, but you cannot find the expected value of anything based only on a sample. If we're sampling from a distribution, then what we mean by that is every observation we are sampling has that distribution. Sample 8 billion values from a binomial distribution with parameters n = 100 and p = 0.4? Then each one of those 8 billion observations can be viewed as a random variable that has a binomial distribution with parameters n = 100 and p = 0.4, and each one of those will eventually take on a value. We're viewing these observations as random variables that eventually take on a value. We are working with known mathematical characteristics about the random variables, not the individual numbers that they've already taken on.

[@gags-villsounds5351]

where can I find the proof for the third step 02:59 ?

[@SobaCha]

Your videos saved me so much, thanks a toooooon!!!!!

[@jbstatistics]

You are very welcome!

[@KrishanSingh-gz9op]

At 00:27, do you mean that expected mean of X1 = Expected mean of X2 = expected mean of X3 .........=expected mean of Xn= μ? And variance of X1=Variance of X2....=variance of Xn= (σ)^2 ?

[@jbstatistics]

"Expected mean" is not a meaningful term here. The expected value of a random variable is its theoretical mean, but we don't speak of expected means. But yes, we are sampling independently from the same distribution, so X_1 through X_n all have the same distribution, implying they have the same theoretical mean and variance.

[@ivoriankoua3916]

So the Mean have a Mean ! what's the Meaning of this Mean ?

[@subtleonset7654]

You are a great teacher, Thanks for the videos

[@jbstatistics]

You are very welcome. And thank you for the compliment!

[@bmgri]

At t=2:40 WHY does the 1/n get squared. That's the part I can't understand. Can you please try to show me how to prove that part slowly using the definition of variance??

[@luciusnguyen2449]

It's a property of the variance. V(aX+b) = a^2V(X). With both a and b are constant

[@ykkung]

Xbar= X(i) if there is only one observation, and is never equal to Mu. Your work of in the video holds true if and only if X(i) is a sample out of the population, and when there are n samples it can prove E(xbar)=Mu and var(xbar)=sq(sigma)/n Not to mention the derivation of var(xbar) is not mathematically correct, as others already pointed out.

[@yaweli2968]

What if we have a huge population and we have sampling distributions of means - say we have about 50 sample means. I guess the mean of the sample means will be the population mean. How will the variance of the sample means look like ? Also, I suspect it’s pop variance if the sample means are large enough.

[@jbstatistics]

As discussed in this video, E(X bar) = mu, and Var(X bar) = sigma^2/n. Those are the true mean and variance of the sample mean. It's best to talk about these things in theoretical terms, because attempts to simplify it with something like 'suppose there are about 50 sample means' just make the situation more complicated. In your example with 50 sample means, the variance of those 50 sample means would be a random variable, with distribution that has a mean equal to sigma^2/n (you didn't give a sample size, so we'll just leave it as n). What the actual variance of those 50 sample means would be is up to the fates, but we know that theoretically Var(X bar) = sigma^2/n, so the variance of a very large set of repeatedly sampled means would be close to this.

[@গোলামমোস্তফা-শ৮থ]

All are good...But I want to know, what does varience of sample mean actually mean? Is there any real example? Like E(x bar) is= mu(pop. Mean)..we can prove it by example..

[@jbstatistics]

You can't prove something like that by example, since all that would show is that it holds for that example. As far as what these things actually mean, I discuss that in other videos. This video is specifically about deriving those formulas. My "Intro to sampling distributions" video, my "Sampling distribution of the Sample Mean" video, etc. The sample mean is a random variable, since its value partially depends on chance (its value would change from sample to sample). As such, it has a probability distribution. E(X bar) is the mean of that probability distribution, and Var(X bar) is the variance of that probability distribution.

[@manupandit7645]

What is the difference between sample variance which is S2 and the variance of the sample mean which is sigma squared / n

[@jbstatistics]

Well, first thing's first. Sigma^2 is the population variance (the true variance of the population from which we are sampling), and sigma is the true standard deviation. S^2 is the sample variance, which is calculated from sample data and estimates the population variance. Sigma^2/n is the true variance of the sampling distribution of the sample mean. The sample mean is a random variable, and like all random variables it has a probability distribution. Sigma^2 /n is the variance of that probability distribution. More info in my intro video on the topic; kzbin.info/www/bejne/p2aTeKOKmauMr6s&ab_channel=jbstatistics

[@medicalstudentscorner3438]

Respected Sir, At 3:18 I didn't get how the variance of the independent observations is equal to the variance of the population. I got mean is the same in both the cases but the variance i didn't get... Could you please help me on that or am I missing something??? Thank you..

[@jbstatistics]

We are sampling n independent observations from a population with mean mu and variance sigma^2. Each of these observations has the same distribution (the distribution of the population from which we are sampling), so they all have a mean of mu and a variance of sigma^2.

[@medicalstudentscorner3438]

Thank you very much sir for clearing up my doubts and additionally I would like to thank you for the efforts you have put into making these videos . It has helped a lot of us and It has also helped us to easily and in an interesting way to understand the statistic part in scientific writing. Much love from Nepal..

[@গোলামমোস্তফা-শ৮থ]

If x1 is just a observation then how E(x1) is equal to population mean mu? Can you explain?

[@jbstatistics]

Each of the X_i are random variables , eventually taking on a value that is a random draw from a distribution with mean mu and variance sigma^2. They all have the same distribution, with a mean of mu and variance sigma^2.

[@zahrariaz8276]

thank you Sir will you please define that what is the theory behind it? i mean why variance of sample mean is always equal to sigma^2/n.

[@jbstatistics]

I'm not sure what you mean. In this video I show just that, that the variance of the sample mean is sigma^2/n. (I take as a given that Var(cX) = c^2Var(X), and the variance of a sum of independent random variables is the sum of their variances.)

[@FrixyFrizzy]

Hey man great video better than my lecturer haha. Was wondering have done a video on the derivation of the ordinary least square estimator?

[@agrimamunjal4041]

Thank you so much.. Please keep making more :)

[@jbstatistics]

You are very welcome. I have a lot of videos in the pipeline, but I've been having a little trouble finding the time to complete them. I'll definitely be adding more in the not-too-distant future. All the best.

[@lukei443]

Exactly what I needed. Thank you!

[@jbstatistics]

You are very welcome!

[@rishabhchopra6418]

Thankyouu SO MUCH! :D Just 2 questions i have: Do you have any video explaining why the variance of the sum is the sum of the variances ? Also, on a related note, do you have any video explaining properties of expected value? Like linearity of expectation? I could not understand it well but surely will if you have a video explaining it :) again, Thanks for everything! :)

[@jbstatistics]

I do not yet have videos on those topics, but I hope to in the future! All the best.

[@matthewkang1946]

The variance of the sum is the sum of the variances only if x1 through xn are independent. If they are not, you cannot apply this property of independence.

[@frankyang9920]

Who can tell me what the software he uses? That you very much I am so curious...

[@jbstatistics]

The background presentation is a Beamer/Latex presentation. I annotate the slides using Skim. I record and edit in Screenflow. All statistical analyses are done in R.

[@frankyang9920]

Wow!! Thank you very much!!!!!

[@sandeepunni9851]

I have a query could someone help me understand this so here X1,X2..,Xn are observations--say height of people from a population of 1000 people and we took a sample of size 100,so here n=100 Now in the calculations we are trying to find variance of each observation which is like finding out how much far away is the observation from the population mean ,so each one should have had a different value ?but here we take all off them have the same value? could somebody please explain this?

[@jbstatistics]

We are sampling n values from a distribution. X_1, X2_, ..., X_n are random variables that take on values once we draw the sample. The distribution of X_1 is simply the distribution of the population from which we are drawing the sample. Similarly for X_2 through X_n (provided that they can be thought of as independent random variables, or in other words, that we are sampling independently from the distribution). When I speak of the mean and variance of X_i, I'm not talking about the actual values that we observe once we draw the sample, I'm talking about the theoretical mean and variance of the random variable. This will simply be the mean and variance of the population (distribution) from which we are sampling.

[@piyushhurpade1371]

@@jbstatistics really thank you so much :) for above one

[@mamma9]

THANKS! Very helpful. // Student at Stockholm School of Economics

[@jbstatistics]

You are very welcome!

[@ibrahimata9934]

What are you doing now ?

[@dinglinduan5058]

Wow, great explanation! Thanks

[@mallatobuckthecanine1750]

I am on the dead-end. Please help me. The 'n' here is a number of samples (n independent observations), NOT the size of a sample. It seems to me that for variance of sampling distribution, sigma squared over n means that sigma squared over "the number of samples," NOT "the size of each (constant through the sample means) sample." But, in the explanation, they somehow intermingled. Please help me understand the proof.

[@jbstatistics]

Here n represents the size of the sample. That is, the number of observations used to calculate the sample mean. I'm assuming that we have n independent observations from a population, and that we will add up those n observations and divide by n to get the sample mean.

[@020dliu8]

jbstatistics in your another video，i saw that when n is large enough, mean of sample mean is equal to population mean. Does this video’s proof process show this thought?

[@jbstatistics]

@@020dliu8 This video shows that the mean of the sample mean is *always* the population mean. You don't need a large sample size for that to be true.

[@020dliu8]

jbstatistics but in another video. You said keep in mind that when n increasing larger, the mean of sample mean is population mean. Now mean of sample mean is equal to population mean doesn’t need large sample size?

[@jbstatistics]

​@@020dliu8 You may have thought I said that, but I very much doubt that I did. (Although I might have been speaking about it in a different context.) If you can point me to the exact video and time, I'm sure I can explain where the confusion lies. The theoretical mean of the sampling distribution of the sample mean is the population mean, and that is not dependent on the sample size. It's not "now mean...", it's always been the case.

[@Kyzcreig]

I love your videos. Question - I'm a bit confused by the notation here:i.imgur.com/rEUUX9v.png What do X1-Xn represent exactly? Are they representing the mean of a given set of samples or are they representing the set of samples themselves? It seems like you use the same notation to mean average, variance, and data set. If the Xi represent data sets it seems strange that you're dividing a dataset by a number, how is that possible. But if they represent variances themselves (which makes sense), I can see how algebraically this has a lot of interesting conclusions. But here's where I'm hung up. If you have 10 samples (X1-X10) of 4 data points each, wouldn't the variance of the sample distribution be algebraically equivalent to taking the variance of those 40 data points? And wouldn't the variance of 40 data points still be the same as the original variance? I suppose I don't understand how taking 10 different means to calculate the variance makes 40 data points more meaningful that taking only 1 mean. But I guess that's the crux of the matter. I guess I'd really like to see some of these operations with actual numbers, if only to clear up confusion about the notation.

[@Kyzcreig]

As a follow up, I get that Xbar represents the mean of the X1-Xn observations i.e. (X1+...+Xn)/n. But how do you take the variance of a mean. Can't you only take a variance of a dataset? And if Xbar is a data set, what is it a data set of? Well that slide seems to indicate it's a data set of (X1+...+Xn)/n but how could you have a data set divided by a number like that. Or is it instead that the Xi-Xn variables are datasets themselves? How can you divide a set of data sets by a variable like that. Am I conveying my point well enough?

[@notthatblin17]

I guess you can think of each xn as infinite number of samples of xn in the sample. that's why each xn has the same mean and variance as the population

[@mightbin]

Thank you very much ,sir. Really helpful!!!!

[@jbstatistics]

You are very welcome!

[@unconditionedrelaxation6726]

i am confused with why 1/n became squared.

[@yaman4863]

thili sadunik It’s a character of the variance, once you drag out a constant from the function the constant becomes squared example: V(3x)=9V(x). I don’t know the ACTUAL theoretical explanation though.

[@RaviShankar-de5kb]

THank you, I saw this in my stats class and felt really dumb... at first I thought "Why isn't Var(x_bar) = sigma^2 ?"

[@jorgemercent2995]

Why is the Var(X1) = sigma square? Why can't Var(X1) = n(p)(1-p)? Anyone, help?

[@jbstatistics]

The variance of the sample proportion: X/n, where n is the sample size and X is the count of the number of individuals with a certain characteristic, is np(1-p). The sample proportion can be thought of as a specific type of mean (the average of n realizations of a binary r.v., taking on the values 0 and 1), but that's not the scenario under discussion in this video.

[@jorgemercent2995]

Thank you!

[@CarlosAugusto-yr3bn]

Incrível, estava procurando isso a um bom tempo, obrigado.

[@TelecomInsights]

great tutorial. well put

[@jbstatistics]

Thanks!

[@capsulecorp6975]

Ive been trying to make sense of var(x1)+var(x2)... = sigma²+sigma²... for three whole days now. can anyone elucidate that for me please?

[@jbstatistics]

When random variables are independent, the variance of their sum is the sum of their variances. This is a well-known mathematical property of random variables, but I don't believe I have a video that gives a detailed explanation. Given that notion, in this video we are sampling independently from a population that has a mean of mu and a variance of sigma squared. X_1 is a random variable representing the first observation, X_2 is a random variable representing the first observation, etc. If we are sampling 2 observations, then Var(X_1 + X_2) = Var(X_1) + Var(X_2) = sigma^2 + sigma^2. (X_1 and X_2 have the same distribution -- the distribution that we are sampling from.) In general, Var(X_1 + X_2 + ... + X_n) = Var(X_1) + Var(X_2) + ... + Var(X_n) = n*sigma^2.

[@adityaprasad465]

Check out 0:32, where he explains that var(Xi) = sigma². In other words, of the terms on the lefthand side you've written is equal to sigma².

[@ycjieyang]

great talk! Thanks!

[@jimsonnevado9783]

why var(x1+x2+x3+...+xn)=varx1+varx2+varx3+...varxn

[@jbstatistics]

If the random variables being summed are independent, then the variance of the sum is the sum of the variances. I don't prove it here, but that it is the case.

[@lynni7595]

You saved me!!!!!

[@jbstatistics]

I'm glad I could help!

[@m3tz13]

Cool , thanks !

[@jbstatistics]

You're welcome!

[@juancarlosherreraburbano194]

lost in 2:26 math.stackexchange.com/questions/1708266/why-square-a-constant-when-determining-variance-of-a-random-variable

[@matinhewing1]

Excellent!

[@jbstatistics]

Thanks!

[@Chicchispirit]

thank you so much

[@jbstatistics]

You are very welcome!

[@2002budokan]

Please enumerate your play lists to obtain a A to Z course easily. Classifying the subjects into play lists is invaluable but play lists dependencies will also be very helpful.

[@jbstatistics]

Thanks for the suggestion, and I'll get to that when I can. For now, note that the videos appear in a reasonable order on jbstatistics.com.

[@mukulyadav2697]

thanks

[@TheProblembaer2]

I love you

[@christopherbuckley2832]

Thank you!!

[@jbstatistics]

You are very welcome!

[@leecherlarry]

i thank you sir. lemme watch more of yu lies

[@celltower1]

Lol why do textbooks have to make the sample Variance look so complicated when its literally just ((1/n)^2)(variance*n)

[@jbstatistics]

I'm not sure what you're getting at here. It's not the sample variance that is ((1/n)^2)(variance*n) = sigma^2/n; that quantity is the theoretical variance of the sampling distribution of the sample mean. And showing why that is the case is meaningful.

[@celltower1]

@@jbstatistics So is Sigma^2/n = var(xbar)?

[@jbstatistics]

@@celltower1 Yes, under the conditions described in this video (that the n observations are independent, essentially).

[@celltower1]

@@jbstatistics Ok I thought var(xbar) was the sample variance