Probability distributions and their properties

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Күн бұрын

Probability distributions, their mathematical treatment, mode, median, mean, and expectation value. (This lecture is part of a series for a course based on Griffiths' Introduction to Quantum Mechanics. The Full playlist is at kzfaq.info?list=...)

Пікірлер: 35

@muddassirghoorun4322 4 жыл бұрын

"CHECK YOUR UNDERSTANDING PART": The graph is a continuous graph and we make use of the integral equation notably: integration of x .rho (x) dx with limit 0 to 1. The mean of x will be 2/3. For the second part, the expected value is 0.8.

@AyushSharma-vl8ey 3 жыл бұрын

thank you for this

@leoulmesfin9384 4 жыл бұрын

The four people that disliked this video were the ones that had zero sexual partners in the probability distribution

@danv8718 2 жыл бұрын

Interesting choice of dataset :)

@dk0r51 11 жыл бұрын

26:35 It would be helpful to have answers to check against. {x} = 2/3 {x^(1/2)} = 4/5

@Debjit625 7 жыл бұрын

I got the same answer

@01rai01 7 жыл бұрын

Me too, same answer

@Oh4Chrissake 6 жыл бұрын

Me too.

@canyuksek4224 5 жыл бұрын

Yeah, I have found the same result, too

@paulayers1111 Жыл бұрын

I got the same

@michaelexman5474 5 жыл бұрын

His visual representation makes this easier to follow at least for me.

@nononnomonohjghdgdshrsrhsjgd 2 жыл бұрын

thanks for the video series! is it shown what kind of distributions the different particles have? I started watching your course on the search of exact and understandable explanation of Ito-integration and measure theory.. now It looks that Ito integral plays no role in QM

@user-xk8uu4gk9l 23 күн бұрын

So is the continous distribution is not a wave funktion. Youd would have to square the wave funktion, and then integrate to get the expctation values for that right?

@jimdogma1537 10 жыл бұрын

Yes, I like these videos and the check your understanding section. Thank you for sharing. The only problem is how are we supposed to check our understanding if there's no answer set to check our results against. Have you posted these somewhere. If not, can you?

@kq6up 10 жыл бұрын

I got the same thing as Anthony below, so I it is probably right.

@debasishraychawdhuri 3 жыл бұрын

Vanishing the first term of the integrations at infinity are somewhat more complicated than simply pointing out one of the parts is zero. What we really have to do is to find the limit at infinity, otherwise, things can go wrong quite easily.

@alannolan3514 4 жыл бұрын

Thanks - jeez, never knew expectation value was the chuffing mean!!!!

@jacobvandijk6525 4 жыл бұрын

Here is an alternative lecture: kzfaq.info/get/bejne/m9mRrMhezNHWmIE.html

@azoooz86 7 жыл бұрын

Thank you for this fabulous video! But at 19:30 how did you conclude that {-x*e^-x} with the upper term is equal to zero? Is not it undefined?

@StuBonham 4 жыл бұрын

Anything to the power 0 is 1. Then 0 * 1 = 0.

@suigetsuhozuki1 4 жыл бұрын

Well, you can write it like this: -x*e^-x = -x/e^x. Now when you plug in infinity you'll get that both numerator and denominator approach infinity, but denominator (e^x) does it "faster" than numerator (x). Recall that exponential function is increasing faster than linear function, so you can imagine it like if denominator reached "greater infinity" than numerator. That way entire fraction is equal to zero. It's kinda weird and maybe even mathematically incorrect to think in that way, but it works :)

@tonyyan4623 4 жыл бұрын

Your method of thinking is totally legit! Another way my friend came up with to think about it is to use the L'Hospital's rule. If we differentiate both the numerator and the denominator in -x/e^x, you get -1/e^x. Then, you can substitute infinity for x to get the correct result.

@shashwat0khurana 3 жыл бұрын

@@suigetsuhozuki1 my maan !!!!

@parthasur6018 2 жыл бұрын

I thought that the probability density function rho(x) at 6.33 would be a bell shaped curve with the peak somewhere between x=0 and infinity 🤔

@a1ang0r85 7 жыл бұрын

I got the answer right, yeah! thanks a lot.However, I don't understand why the probability density can be presented in this way? 9:45 also why expected value of the function can be presented in this way? 21:15

@239manognya2 Жыл бұрын

I hope you found the answer,if yes please explain...

@aidansgarlato9347 6 жыл бұрын

I know this was not intended I'm just more of a pure math mathmitician so I recognized the gamma function when I saw it so for the last integral the question simply boiled down too x! as that is the definition of the gamma function making it be equal too 2! which is simply equal to 2.

@shashwat0khurana 3 жыл бұрын

r/bon apple tea