For more information on econometrics and Bayesian statistics, see: ben-lambert.com/
Пікірлер: 37
@mayankjangid15433 жыл бұрын
Finally got the intuitive reasoning of the inequality ! Thank you so much sir!
@baqerjawadal-lawati84732 жыл бұрын
Simplicity is the ultimate sophistication. Thank you for showing that.
@Issara867 жыл бұрын
All of a sudden utility maximization under uncertainty makes so much more sense now. Thank you good sir.
@prestonhansen497 жыл бұрын
Thanks a bunch for the video, this is a lot easier to follow and less notation heavy than the proofs from my course!
@solfeinberg4375 жыл бұрын
So clear, such an efficient proof - thank you for informing me and respecting my time and intelligence. Bravo!
@jacobschuster4857 Жыл бұрын
Such a clean and intuitive way of proving this. Thanks!
@ericarcherman99559 ай бұрын
Great video here! Thanks!
@user-tl8xb8ks6s5 жыл бұрын
Doesn't L[E(x)]=g(E(x)) only hold if the x is the particular point where g(x) and L(x) meet? Generally, L[E(x)]=
@ericzhang76585 жыл бұрын
L[E(x)] is defined such that it is tangent to the function g at g[E(x)]
@marofe4 жыл бұрын
Actually it is correct! The proof relies on the convexity inequality, but you also can prove by a tangent plane if you choose L(x)=g(E(x))+g'(E(x))(x-E(x)). thus, as g(x)>=L(x) for all x we got that E(g(x)) >= g(E(x)) for all x.
@tristanseow29703 жыл бұрын
L[E(X)] and g(E(X)) are both constants, for a random variable X, thus they should be equal to each other
@dinaashraf91995 жыл бұрын
That's really informative and easy to be understood. Thank you.
@emrglr6963 Жыл бұрын
great way of proving this, thank you!
@pinkability55472 жыл бұрын
Brilliant. It is much easier than I thought :))
@TheRyry979 жыл бұрын
This is great. Never seen Jensen's proved in this manner before.
@deductionism9 жыл бұрын
Ryan Tamburrino It is proven in exactly the same manner at the book A First Look at Rigorous Probability Theory by Jeffrey Rosenthal
@samuelbarham84834 жыл бұрын
Oh that's such a beautiful proof.
@andrewrochman88594 жыл бұрын
how cool it is i can actually find minimum values of trigonometric functions in a triangle by using jensons inequality
@yaminireddy51573 жыл бұрын
Yes, you can but as trig functions are periodic, and triangle angles have a condition like at max only one obtuse angle/right angle, you should be very careful as some angles may land on the convex part while others concave!
@krstev29Ай бұрын
Do you know how complex is the induction proof?
@activision41704 ай бұрын
Would this also imply the reverse is true for a concave function? I.e., g(E(x)) > E[g(x)] ?
@lazarbeic11313 жыл бұрын
Very smooth, great!
@vidajohn21997 жыл бұрын
Thank you. This is very clear.
@iamoo6 жыл бұрын
Thank you! Couldn't understand explanations from the Blitzstein's book..
@zihadazad7 жыл бұрын
Nassim Nicholas Taleb Brought me here
@sucim5 жыл бұрын
How did that come about?
@hanjunkim92986 жыл бұрын
Thank you so much!
@barosderiche6455 жыл бұрын
Thanks but not rigorous. A convex function is not necessarily differentiable. Thus there may not exist a tangent line at all.
@giorgioyoung1025 жыл бұрын
it's differentiable at all but finitely many points, so not such a big lie
@user-yc3yj5nn3z7 жыл бұрын
thank you~
@dorian96364 жыл бұрын
awesome
@Tomvik19754 жыл бұрын
6
@kaidongwei44065 жыл бұрын
doesnt make sense
@kaidongwei44065 жыл бұрын
I will use it anyway since all of u say it is right
@kaidongwei44065 жыл бұрын
bad proof
@itstotallydope5 жыл бұрын
@@kaidongwei4406 HAHAHHA which part do u not understand