Deriving the Normal Distribution Probability Density Function Formula

Рет қаралды 21,791

MolloyMaths

3 жыл бұрын

Пікірлер: 56

@jeevanjose6986 Жыл бұрын

Brilliant explanation! Love how you took care of any questions we would potentially in between the steps. Subscribed! 🙏

@molloymaths1092 Жыл бұрын

Thanks and your welcome!

@marcushendriksen8415 Жыл бұрын

Great video! I particularly liked how you took the time to highlight the relationship between lambda and the spread.

@molloymaths1092 Жыл бұрын

Thanks. Glad you liked it.

@whistlinghouse6488 2 жыл бұрын

Excellent explanation. Great job. I am particularly impressed that you explained even the minutest details. Thanks. Can you post similar videos for other distributions as well. For example, Beta, Weibull, Poison etc.

@molloymaths1092 2 жыл бұрын

Thanks. Haven't done those yet but will have a look at doing them at some stage.

@vikramanbaburaj525 4 ай бұрын

Will be helpful for those want to know deep about the topic.

@wahidin5075 Жыл бұрын

Very great video sir! I am very happy get your explanation about deriving the pdf of normal distribution

@molloymaths1092 Жыл бұрын

You're welcome

@kamranabbas2438 Жыл бұрын

Excellent video, especially when you piece it together with derivation of the Gaussian Integral. Thank you very much!

@molloymaths1092 Жыл бұрын

Your welcome. Glad you liked it.

@Rondon0905 2 жыл бұрын

thank you so much. i remember watching this a while ago and now I needed it. happy you had it on youtube still. I'll prefer to fix k = -1 first to make the derivation easier. it follows sigma^2 = 1/2. and from that we can apply lineartransformations.

@molloymaths1092 2 жыл бұрын

Your welcome. Glad it helps.

@KryDu-lv3jk Ай бұрын

Exactly what I am seeking . Thanks

@molloymaths1092 Ай бұрын

Great. Glad it helped.

@akilanramesh725 4 ай бұрын

Haven't seen any video or full proper explanation about this derivation, from a student in need for his school main project thanks alot and im definitely sourcing you for my project(don't worry lol).

@molloymaths1092 4 ай бұрын

Great. Glad it helps. Good luck with your project.

@xaqaniqasimov7525 Жыл бұрын

Thank you very much,perfect explanation,and now I can feel comfortable...

@molloymaths1092 Жыл бұрын

Thanks. Glad it helps.

@johannaw2031 2 жыл бұрын

Helpful. Thank you for going through every step.

@molloymaths1092 2 жыл бұрын

Your welcome.

@johannaw2031 2 жыл бұрын

@@molloymaths1092 Hi, I was presented this density function in my finance course as a part of the Black-Scholes formula (which I guess you have knowledge of). Im trying to grasp where this actually comes from and your video has helped a lot. However, I have a couple of questions and would be grateful if you could try to help me out. 1. How can you just make the assumption that y is 0 and what kind of implications does this really have? I mean, if y is 0 then graphically you just have an x-value going right and left? 2. How do you actually know that you can change g(x) to a function of the form e^kx^2. Is this a result of some kind of trial and error? I do know that e is "important" but why for example dont you try a function like g(x) = 25 - 3x/4 + x^2 ? To my lack of mathematical insight it seems like you took it out of thin air. Thank you and have a nice day! Best regards, Johanna

@molloymaths1092 2 жыл бұрын

@@johannaw2031 Hi. Thanks for your comment. You can let y=0 since you are doing it on both sides of the equation. Also it doesn't change the probabilities as r doesn't change. There is a certain amount of trial and error as well as intuition when substituting e^kx^2. It does work as I have shown that in the video. Thanks again for your comment.

@wqltr1822 Жыл бұрын

I wonder if there are variants of the normal distribution, by this I mean continious functions who satisfy the original assumptions and ehen integrated from neg inf to pos inf equals one. This could be explored by taking the line g(x)g(y)=g(sqrt(x^2 + y^2)) and trying to find other solutions for g. Or maybe it can be proved the only a certain family kf functions exist, idk.

@rachmatsaleh4219 2 жыл бұрын

Excellent and thank you so much

@molloymaths1092 2 жыл бұрын

Your welcome.

@Amsemsora Жыл бұрын

Thank you!

@molloymaths1092 Жыл бұрын

Your welcome!

@manikandank7445 Жыл бұрын

Excellent explanation Sir👏👏, thank you

@molloymaths1092 Жыл бұрын

Your welcome!

@vedantjhawar7553 2 жыл бұрын

Quick question. After rewriting the function based on f(y) where y=0 is a constant, everything else flows smoothly. I dont understand, however, how y itself can be regarded as static without the derivation restricting calculations of other y-values. Do let me know if you would like me to clarify my question. Thank you.

@molloymaths1092 2 жыл бұрын

Not sur exactly what you mean!

@abdul-kadersouid515 Жыл бұрын

Outstanding, clear and precise. How about if you add examples to the use of the CDF? With many thanks and appreciations

@molloymaths1092 Жыл бұрын

Thanks for your comment.

@madmorto2610 Жыл бұрын

Excellent video!

@molloymaths1092 Жыл бұрын

Thanks!

@madmorto2610 Жыл бұрын

@@molloymaths1092 were there any assumptions made about the spread of darts initially? And would other functions exist that satisfy the functional equation you derived? Ones that would produce other, just as valid probability density functions?

@molloymaths1092 Жыл бұрын

@@madmorto2610 I didn't make any assumptions about the spread of the darts. Sorry but I'm not sure of any other functions that can be derived from the functional equation.

@madmorto2610 Жыл бұрын

@@molloymaths1092 ok, I think there is an implicit assumption made about the spread of the darts when choosing to use exp function.

@chengchuanliu7839 2 жыл бұрын

Fantastic Video!!! I love the way you make the explanation! A quick question at 31:10 I don't quite understand why you consider x goes to infinity and the red-circle-part becomes null. Surely if we plug in 0, it is 0 as well. But if we plug in 1, it has value not equal to 0. Thanks!

@molloymaths1092 2 жыл бұрын

e to the power of minus (Pi lambda squared x squared) is 1/(Pi lambda squared x squared) which becomes zero if x goes to infinity.

@hiuyingchoy5399 Жыл бұрын

@@molloymaths1092 i understand its the denominator. But why would x goes to infinity in this case?

@molloymaths1092 Жыл бұрын

@@hiuyingchoy5399 We are integrating from +/- infinity

@pjakobsen Жыл бұрын

What is the symbol you call "ORR" or "OHR" that looks like an r? Is this a greek letter ?

@molloymaths1092 Жыл бұрын

It's just an r

@onepunchman8721 Жыл бұрын

Do you multiply f(x) with f(y) because you want to express AND?

@molloymaths1092 Жыл бұрын

yes.

@harekrishnapradhan5049 4 ай бұрын

I couldn't understand If w(x)=lambda.f(x) only when y=0 then after that all the steps involving y are just identities like 1=1 Explain?

@molloymaths1092 4 ай бұрын

Where?

@xaxion_faza2453 Жыл бұрын

A very elegant video. A quick question, why does w(r) = f(x).f(y) ?

@molloymaths1092 Жыл бұрын

Thanks. The probability of the dart landing in the x direction is independent of the probability of the dart landing in the y direction. When two events are independent you multiply their probabilities.

@user-oc4gu9kb3p Жыл бұрын

Can you please tell the reference?

@molloymaths1092 Жыл бұрын

Not sure what you mean

@user-oc4gu9kb3p Жыл бұрын

@@molloymaths1092 If you have learned this from a specific book, tell me its name if you could remember. thank you for replying.

@elijahhallbasketball Жыл бұрын

Enjoying the video! But at 10:43 when you derive that [2]: w(x) = L*f(x), we can pair this nicely with the fact directly above [1]: w[sqrt(x^2 + y^2)] = f(x) * f(y) to yield: By substituting [2] into the LHS of [1] L * f(sqrt(x^2 + y^2)) = f(x) * f(y) We can then divide both sides by L^2 to obtain the result you achieve around 12:26 L* f(sqrt(x^2 + y^2)) / L^2 = f(x) * f(y) / L^2 => f(sqrt(x^2 + y^2)) / L = f(x) / L * f(y) / L I think this step would probably have been a lot quicker than what you used, but perhaps there is a reason you followed the steps you did? Again, enjoying the vid!

@molloymaths1092 Жыл бұрын

I think that there are a number of ways of getting through this derivation. I have just given one version. Thanks for watching and for your interesting comment.