Mean & Variance of the Exponential

[@mikhailkilianovski8024]

For me, it is more intuitive to find P(min(X1, X2)

[@ShinCharles]

That was a really informative video. I felt like I just reviewed my calculus math.

[@MinhNguyen-gg2iu]

Thanks. It's very helpful reviewing the calculus thing!

[@عبداللهاللهيب]

this is perfection . greet explaining keep it up

[@عبداللهاللهيب]

and by the way you just explained two concepts subjects (exponential and order statics ) on 15 min' s on other uni's this can be a two lectures 90min's each

[@farukben]

Great work! Thanks.

[@randalllionelkharkrang4047]

why isP[ max(x1,x2,x3)] equivalent to P[x1

[@randalllionelkharkrang4047]

oops , sorry , i finally got it. hehe. if max

[@cyenkwang]

Exponential Var(X)=1/lambda^2

[@maggem123]

Good one. Well explained.

[@raginibhayana8305]

you are really good : )

[@mariafernandamelo6861]

what if I wanted teh E[ X^3]?

[@jonahberrong6761]

can we just use theta for the mean. lol why are we making it harder than it needs to be

[@lijerry9396]

can you calculate lamda out? by integrating fx from 0 to infinity

[@justpaulo]

No. No matter what λ is, exp(-λx) is always zero @ x=∞, which means that the CDF = 1- exp(-λx) is always =1 @ x=∞ independently of λ.

[@2009worstyearever]

why did t turn into negative t suddenly ? where did the negative appear for the -e from ?

[@jagdishramakrishnan8190]

+2009worstyearever I'm assuming you're referring to the result in part a. Integrating exp(a*t) with respect to t gives us (1/a)*exp(a*t). The pdf of an exponential is lambda*exp(-1*lambda*t). So, when integrating the pdf, we would get -1*exp(-1*lambda*t). Hope that helps.

[@sathithvindika9871]

Great

[@AmbroseOdhiambo28]

Good example there but the board was reflecting a lot of light, I strained a lot while following the tutorial

[@footballonfire2704]

where is the lamda in integer of variance? wkwkwk :D