Moment Generating Function #2 (Continuous Distribution)

[fenekfenekk]

seek, and ye shall find. Thank you so much for making this video !!!

[betterkenya]

step 4 kida thow me from the pan to the fierce fire. where did you get the constrain that t-µ‹0?

[stepbil]

@betterkenya dude. Listen carefully to what i said in the vid. (t-λ) must be less then 0 in order for the integral to converge and not explode in our face. To convince urself let's assume that t-λ is a positive number say 2. And now evaluate this integral from 0 to infinity. U will notice that you cannot evaluate because the integral will blow out on the upper limit of integration (here infinity). hope it makes more sense now.

[Burrowsnakes]

typo at property #8: should be 5/5-3 not 5/5-2

[stepbil]

@Burrowsnakes . Well done Burrowsnakes! I thought I will wait an eternity for someone to spot it ;-).

[zisongxD]

YOU SAVED MY DAY

[stepbil]

You will find the key to this mystery if you watch #1 in the series.

[imkb010]

Where did the expectation of e^tx come from? I realize that the MGF comes from it, but WHY? It is still mysterious and unintuitive to me why it works.

[stepbil]

I think you have only focussed on the numerator. But what would happen to the denominator if we allow (t-λ) ≤ 0 and say let t-λ=0. We would end up with 0 in the denominator and we don't like that do we ;-)?

[apanapane]

Just realized that it wouldn't matter that exp(x(t-λ) would converge to 1 when (t-λ) = 0 since the next step anyway leads to e^0.

So the question is really, shouldn't the integral only be defined when (t-λ) ≤ 0 instead of <0 ?

[apanapane]

Nice video. Thank you for taking the time to make this.

One question though. You say at around 6:50 that exp(x(t-λ) wouldn't converge unless (t-λ)<0.

However, if (t-λ) = 0, the expression would converge to 1. Does this provide a problem? Or is my reasoning faulty.

Cheers