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
+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.
For me, it is more intuitive to find P(min(X1, X2)
That was a really informative video. I felt like I just reviewed my calculus math.
this is perfection . greet explaining keep it up
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
why isP[ max(x1,x2,x3)] equivalent to P[x1
oops , sorry , i finally got it. hehe. if max
Thanks. It's very helpful reviewing the calculus thing!
what if I wanted teh E[ X^3]?
can we just use theta for the mean. lol why are we making it harder than it needs to be
Great work! Thanks.
why did t turn into negative t suddenly ? where did the negative appear for the -e from ?
+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.
Exponential Var(X)=1/lambda^2
can you calculate lamda out? by integrating fx from 0 to infinity
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.
you are really good : )
Good one. Well explained.
where is the lamda in integer of variance? wkwkwk :D
Good example there but the board was reflecting a lot of light, I strained a lot while following the tutorial
Great