The memoryless property of the exponential function, implies you could infinitely zoom in the the graph of the distribution, and it will always look the same. It is very intuitive when you think about it in this way.

Gordon Freeman’s favorite distribution

beautiful math that I will forget next semester but the feeling would remain

Great series!  Thanks!

A less murdery radioactive decay example is NASA's use of  plutonium-238 in  Radioisotope thermoelectric generators (RTGs) - decades of electrical power for space probes traveling far beyond where solar panels become impractical (the Voyager RTGs have lasted almost 50 years so far, and also help keep electronics warm enough to function). 🙂

but the probability of any single t is 0 right because is X continuous?

2:35 It might actually be a problem of the power supply instead of the light bulbs...

7:54 there should be lambda in place of t in the calculation of definite integral between b and a

12:36 a particle flew out of his mouth lol

can. we get the notes of this complete lecture series, might help to revise quickly

Average period of Valve releasing a major game is 3 years. What is the probability that they drop Half Life 3 this year?

there is an error in 7:25 should be e^(-lamba*a) - e^(-lamba*b)

Video published 44s ago but comments from a month ago?

7:50 there will be lambda in exponent not t

I don't like that distribution formula just pops out of nowhere  - it can be easily derived using  two assumptions: 1. at t=0 probability of decay equals to λ*dt for any sufficiently small dt. 2. at t = T provided that no decay has occur for 0 < t < T, the probability of decay for T<t<T+dt is λ*dt (the property of memorylessness, "atoms do not age" ). Lets calculate P(t): using Bayes' formula, probability of decay for  t, t+dt equals to (1 - Integral(0, t, P(t), dt)) * λ * dt  (the first term is the probability that atom survives until time t). Therefore P(t) = (λ*(1-Integral(0, t, P(t), dt)))). Take the derivative of both parts: P'(t) = -λ*P(t). Then integrate using separation of variables dP/P= -λ*dt => ln(P) = Const - λ*t. Then calculate Const: ln(P(0))=ln(λ)=Const. => P(t) = exp(ln(λ)-λ*t) = λ*exp(-λ*t)

Thanks for great explanation. Would you please tell me how did you creat video? How is it possible by writing such, seems on a glass! we can see both you and writing, it's amazing, any AI used?

i think the definition in the first part is wrong. He says probability that T=t is lambda*exponential. Which is wrong. It is the pdf.

🍿:trophy-yellow-smiling:😁🥳👍👍💪

how do you choose the distribution for a given scenario? i understand as this one is time related, but is it always exponential when time involves?

we should write f(t) and not P(T=t) for a probability density, because P(T=t) = 0 if T is a continous random variable ;)