Deriving the Mean and Variance of a Continuous Probability Distribution

[@Manny123-y3j]

I am blown away by how well you explain things. Your videos are simple and to the point yet so incredibly clear. I appreciate your work so much. Thank you!

[@uvindujayasinghe6195]

You have really played a large role in helping me understand probability and statistics which is one of my weaknesses in maths. Please keep doing what you're doing.

[@jbstatistics]

+Uvindu Jayasinghe Thanks for letting me know Uvindu. I'm very glad I could be of help! All the best.

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[@ely4529]

The best channel for statistics, thank you so much for the effort!

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[@alexandertam2542]

Damn bro your videos are mad concise, i learned this real quick. Appreciate the help man

[@jbstatistics]

You are very welcome Alexander. I'm glad to be of help!

[@jiayi9603]

Same here

[@Lexyvil]

Thank you very much for being clear. My university taught calculating E(X) for Discrete Random Variables, but did not go into E(X) for Continuous Random Variables, which this video explains very well.

[@jonathanlindberg1426]

I have my econometrics midterm tomorrow morning and wow, this video was insanely helpful as my professor doesn't provide textbook, worked examples or powerpoints. Thank you!

[@actionjessie]

OMG you make this so easy to understand and simplified! Much appreciated

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[@jbstatistics]

I'm glad to be of help!

[@Banditxam4]

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[@jbstatistics]

I'm glad to be of help!

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Hi! I'm student of Yonsei University in Korea, and my major is Applied Statistics. Your videos are really helpful for my study :) Thank you🙂

[@jbstatistics]

You are very welcome! I'm glad to be of help!

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mid-term tests next week, You are my hero during this revision :)) Thank you

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[@GoodLuckForever-wi9kb]

Thanks , best method to explain such difficult concepts.

[@darthgeorge2]

Amazing, clear and thorough tutorials, you've helped me fully understand Continuous Probability Distribution's. Thank you!

[@jbstatistics]

You are very welcome George! Best of luck in your studies.

[@tasneemsakifibnealam949]

Your videos are on point and easily understandable. Thank you for helping students like me.

[@tofubtofub]

Really dont know how to express my gratitude to you

[@vemuyaswanth803]

BRO! Thnx! Forever in your debt!

[@josephwheelerton]

Why is the definition of the E(X) equal to that integral in particular? The pdf f(x) value doesn't actually represent the probability of the x value it corresponds to, so then why is the definition of E(X) almost identical to the definition of E(X) for a discrete random variable? I am having some trouble understanding what I am summing with the integral.

[@birindergiddey7734]

See this video: kzbin.info/www/bejne/jKrOaXuHhMief6c

[@jbstatistics]

The video that you commented on previously ("Finding probabilities and percentiles for a continuous probability distribution") works through an example of finding the median (and possibly another percentile). Your problem is very similar to the one in that video, the main difference simply being that the function being integrated is not the same.

[@sridhargampa3661]

Excellent way of teaching !! subscribed rightaway!! Thank you so much!!

[@bensmyth450]

This makes so much sense and so beautiful thank you

[@pantrucaxd]

Thank fou very much for the video! great explanation!! exactly what I was looking for

[@jbstatistics]

You are welcome! I'm glad to be of help.

[@sathyanarayanankulasekaran5928]

eye opener....real thanks for your efforts..

[@ren222]

You are my life saver this course

[@jbstatistics]

I'm glad to be of help!

[@lidiatoscano5334]

I really like your explanations. Very understandable. Thank you!

[@russellkilgour]

Thanks JB, im now a belieber in your omniscience!

[@jbstatistics]

You are very welcome Russell! It's always good to have a few beliebers.

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[@hasibahmad297]

Though I am watching now, these videos were published 11 years ago. Feeling like I am thousand years behind the civilization.

[@jbstatistics]

Or maybe I was 11 years ahead :) I tried to build them to stand the test of time.

[@hasibahmad297]

@@jbstatistics I really wish you were my one to one stat teacher. I have been sinking in the ocean of stats 😞. Idk how many times I come back to this channel to watch the videos over and over again whenever I feel like I am struggling to understand a concept. Cannot thank you more.

[@jbstatistics]

@@hasibahmad297 I'm very glad to be of help. I'm still (kinda) your one on one teacher :)

[@funfacts-hatari]

You're a life saver

[@jbstatistics]

You are very welcome!

[@TruthBeTold.]

Well done sir! Thank you for your contribution

[@borenxu8178]

Great video, really helpful refresh rusty memory, thank you so much! but just one question, at 5:06, should the integration equation be using f(x) instead of F(x), can you please clarify. Thank you

[@jbstatistics]

It is f(x), and that's what I wrote. But it can be a little tough to distinguish f(x) from F(x) in handwriting. Cheers.

[@jack950911]

Thanks for your video, it help me save 3days to understanding what lecturer teaching in college, haha

[@logicdedady1989]

It's nice for all students

[@akjm9177]

You saved my maths grade

[@jbstatistics]

I'm glad to be of help.

[@christiechen193]

omg, this is super helpful!! thank you so much.

[@KennedyOmoniyi_Agbagi]

Your a life saver

[@adiknoxrider1]

Amazing video! Explained it very well!

[@jbstatistics]

Thanks!

[@aishashahzadi7119]

Almost got it. I have a confusion here.while you were deriving expected value,1/60(x^5/5)... How did we get 5. I have been stuck here for quite sometime and no one is able to explain so far

[@jbstatistics]

That's a basic integration: the integral of x^4 is x^5/5 (+c). In other words, the derivative of x^5/5 is x^4.

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[@tompage8438]

you have reduced the chances of me failing my 50% exam thank you sir i wish i could kiss you forehead

[@samo6391]

I do see some formulas where the integral of f(x) is multiplied by 1/b-a where a and b are the interval. what's the difference?

[@jbstatistics]

I really don't know what you're asking. If you rephrase it I might understand.

[@গোলামমোস্তফা-শ৮থ]

But I want to know how did you find the equation of the expectation of Continuous random variable. Why this equation is equal to mean or expectation?

[@jbstatistics]

That's the mean of a continuous random variable by definition. It makes sense to weight each value by the likelihood of it occurring, and the result is the value that the sample mean tends to as the sample size goes to infinity, but the formula is the formula by definition.

[@chimanruler15]

You're amazing! Thank you sooooo much! =D

[@JustinQuiksilver]

Hello jbstatistics. The limits of your pdf are x greater than or equal to 2 and x less than or equal to 4. Suppose the limits were greater than and less than without the equal, would you still integrate using the exact limits of 2 and 4?

[@jbstatistics]

Yes. If X is a continuous random variable and c is any constant, then P(X=c) = 0. (Probabilities are areas under the curve, and the area above any point is 0.) So P(X

[@johnyumbezuze1958]

When finding (e^2) why is the power suddenly being raised to 5 from 4

[@geraldidzakwan8666]

You saved my life dude

[@maqboolahmad2928]

Mind blowing sir

[@nikosroom1913]

At 4:50 you said F(x), which confused me because I thought you meant the cdf... otherwise great video, but it's definitely the pdf here and not the cdf!

[@jbstatistics]

Yes, it's the pdf. I wrote a lower case f, and it looks like a lower case f to me. On the line immediately following, in the integrand I replaced "f(x)" with 1/60 x^3, which is given as f(x) (between 2 and 4) above, so I don't think there can be much confusion for long. In handwriting, it can be tough to distinguish a lower case f from an upper case F sometimes, but my upper case F would never cross the vertical line like that.

[@sohail2318]

supremely excellent sir

[@jbstatistics]

Thanks!

[@rimshazaidi2375]

does this means that formula foe the variance is same for discrete and continuous variable?

[@jbstatistics]

Yes, in the sense that for any random variable X, Var(X) = E[(X - mu)^2].

[@almawamulo787]

Thank you, Thank you, Thank you.

[@jbstatistics]

(You are very welcome)^3

[@shreyasdamle4755]

Nice Work Sir!

[@simonhuisman8866]

If we want to calculate the standard deviation. Then we use the variance and it is √0.266 right?

[@jbstatistics]

Yes.

[@bewanthawijesinghe4266]

superb explanation !!!

[@jbstatistics]

Thanks!

[@ozanertek4238]

these videos are very helpful :) thanks a lot :)))

[@roadtoad3484]

When I intergrate from 4 to 248/75, I get 0.56853? Why is it not 50% or even really close to 50 percent?

[@jbstatistics]

You found the area to the left of the mean. The area to the left of the *median* is 0.50, but the area to the left of the mean can be pretty much anything, depending on the shape of the distribution. If it's a symmetric distribution, then the mean and median are equal, but this distribution is definitely not symmetric.

[@roadtoad3484]

@@jbstatistics Thank you for the clarification. I am taking an econometrics class at UCSD and This channel is really helpful.

[@Stekaren]

thank you so much sir!

[@jbstatistics]

You are very welcome!

[@11spongebob]

Anything about marginal and conditional distributions?

[@axadify]

Thanks a lot man, this was really helpful :)

[@jbstatistics]

I'm glad I could help!

[@zogalz5598]

Thanks for the very useful video

[@jbstatistics]

You are very welcome.

[@ma70vrsn2jz]

ah yes another fine ass saving by Mr. jb.

[@crapzone1]

why do u increase the power of x when taking out 1/160 out of the integral? at (1.56)

[@crapzone1]

+jbstatistics damn it, so obvious and thank u tank u thank u , thank you for the videos.

[@jbstatistics]

+hardeep josen You are welcome!

[@sanpedroryanjoshuag.5645]

what should i do if the mean i get is no inside the limits?

[@buttercup2597]

Thanks for explaining so beautifuly

[@karimkhan1312]

i wish i can see u some day and learn from u

[@lientranthikim2845]

sorry, could you please explain why variance of x = 0,266. ? I tried lots of time but I just got the minus sign before

[@jbstatistics]

I'm not sure what you are asking. Finding the variance took several minutes in the video, and I don't see how I could flesh out the explanation more than what is given in the video. The final calculation is correct, and if you're not getting the right answer there I'm not sure where you are going wrong.

[@LienTran-sy7ed]

jbstatistics sorry for my mistake. I mean how did you calculate 56/5 - (248/75)^2=1496/5625. My result was -216... Idk why 🤔😞

[@jbstatistics]

I'm not sure what you are doing wrong. Try this in your calculator: (56/5)-(248/75)^2.

[@Me050292]

THANK YOU I LOVE YOU

[@motormusique]

I know a Canadian when I hear one. Also, thank you.

[@user-xq9il2mp1q]

Thank you very much, you helped me

[@garyanfimau7203]

In my understanding, what was done here is compute, not derive. I expected the explanation behind the formulas for the mean and variance of the distribution.

[@jbstatistics]

Those are the *definitions* of the mean and the variance of a continuous probability distribution. Any explanation would simply be describing why those are meaningful quantities to know. I took the definitions of the mean and variance of a continuous random variable, and, using a series of logical and mathematical steps, figured out what the values are for this particular distribution.

[@actionjessie]

Ive been given an exponential function and i need to find Q1 Q2 and Q3, can you give me a hint? all i have is 2exp-2x for x

[@karimkhan1312]

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[@Jakim-James]

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[@jbstatistics]

Thanks!

[@InstansVisio]

i don't understand at 2:02 how do you get the 1/60 . (x^5/5)

[@InstansVisio]

+Killan TRAPSA never mind i understood!! man u are a master!

[@ozanertek4238]

because x.f(x). so x .x^3 = intagrate x^4=> x^5/5 . i hope i could help you.

[@haiqqalosman2644]

Is it same formula with moment about mean?

[@jbstatistics]

The mean is the first moment about the origin. The variance is the second moment about the mean.

[@SAS_OP]

Thank you so much

[@vansikasingh3228]

Thank you so much!

[@hima3x46]

Brilliant ❤️

[@ssmputri]

you save my day :)

[@soumyajitsarkar2372]

Thank you, Sir :))

[@jbstatistics]

You are very welcome!

[@cliffordochieng3596]

man God bless you

[@jbstatistics]

I'm happy to have the blessing!

[@syedrizvi2687]

Thank you!

[@sim4552]

Intuitively, I would expect E(X^2) = Integral(x^2*f(x^2)*dx)....

[@jbstatistics]

I get a (very reasonable) comment like this from time to time on a number of different videos. Last week I gave this reply to a similar question: We could find E(g(X)) by using E(g(X)) = sum g(x)f(g(x)), where the summation is over all possible values of g(x) and f(g(x)) = P(g(X) = g(x)). But the law of the unconscious statistician tells us that E(g(X)) also equals sum g(x)p(x), where the summation is over all possible values of x and p(x) = P(X=x). This saves us from having to find the distribution of g(X). (In your comment, you'd also need dx^2.)

[@sim4552]

jbstatistics thanks! I did think of dx^2 when I was writing it out. Your explanation makes sense!

[@huesOfEverything]

How is this derivation?

[@odeyemiadejoke750]

God bless you

[@stutikumar4384]

thank you thank you thank you

[@emmanuelmtali1594]

Standard Deviation?

[@jbstatistics]

The standard deviation is by definition the (positive) square root of the variance. So to find the standard deviation we first find the variance then take the square root.

[@ericvanwageningen4340]

actually, you do have the right idea, but as someone who has passed the first two actuarial exams, Probability and Financial Mathematics, I can tell you that I learned on the former exam that, in order for a Probability Density Function (PDF) to be valid, it's lower limit must yield a value of 0 in the Cumulative Distribution Function (CDF) and the upper limit must yield a value of 1. Instead, when we integrate the CDF to x^4/240, we get a LL of 16/240 and 256/240 for our UL, when in fact, the LL should be 0 and the UL should be (240)^(1/4). Other than that, you do have the correct idea, as far as how to integrate is concerned. Or were you just arbitrarily picking an example just to show how it'd be done, ignoring the validity factor?

[@jbstatistics]

As someone with a doctorate in statistics, I understand everything that you're saying, but I suggest you check your math again. Yes, the lower limit of the CDF must be 0, and the upper 1, as they are for my chosen function. First note that I give pdf and do not discuss the CDF in this video. The pdf integrates to 1. Int_2^4 x^3/60 dx = 1. The CDF is not discussed in the video, but is 0 for x < 2, 1/240 (x^4-16) for 2 4. There's always the chance that I'd make a mistake somewhere, but I don't just pick functions willy nilly and ignore the very basics of probability and statistics.

[@ericvanwageningen4340]

oh I see what you're saying. Sorry, is this like a mixed distribution or something? I remember those from Exam P as well. And btw, I didn't mean to sound like an arrogant know-it-all by alluding to the fact that I passed the first two actuarial exams; I meant to sound like an anomaly, if you will, given the rarity of people who even know what an actuary IS, much less those who have passed one or more of the exams! Sorry, I didn't mean to sound condescending; but I could have sworn that the LL always had to result in 0 when plugging into the CDF and the UL had to result in 1, which, ordinarily, would be true, say, for f(x) = x^2/9, this integrates to a CDF of x^3/27, for which our parameters are 0

[@ericvanwageningen4340]

and to be honest, I'm actually quite used to people not knowing what an actuary is, so, it's really not a bother being asked what it is. All I'm saying is I wanted to emphasize the fact that it's not everyday that you meet someone who is studying to become an actuary, that's all.

[@jbstatistics]

The distribution I use in this video is a basic continuous probability distribution. Its support is [2,4]. The random variable X can't take on values between 0 and 2, which is confusing you here. It's no different from your x^2/9 for 0 < x < 3 example, it's just a different function with different support. The CDF of my distribution does indeed have a lower limit of 0, an upper limit of 1, and is non-decreasing, as any CDF must. Feel free to plot it and see what it looks like.

[@ericvanwageningen4340]

okay. yes, I believe you; in fact, if you were to take the PDF of f(x) = 4x^3/175, you see that's a valid PDF for which 3

[@halilkazik7816]

Thank you

[@SaurabhKanawade]

Intigrate how calculate in3rd step

[@ferny4603]

Thanks bro

[@huesOfEverything]

This is not derivation... this is solving based on formula. Misleading title

[@MrMarkgyuro]

why x*f(x) 🥺?

[@marcoventura9451]

I think that you should watch this video on E(x) of a discrete probability distribution to catch the analogy: x*p(x) is the discrete "version" of x*f(x) : kzbin.info/www/bejne/jKrOaXuHhMief6c Moreover you should be familiar with integration, i.e. how the summation becomes an integral.

[@MrMarkgyuro]

@@marcoventura9451 thanks for your help!

[@dunethchadeera5838]

❤️❤️❤️