Help me show this

Y

yeye12

New member
Joined
Oct 3, 2014
Messages
3
Please help me show this

Hiii, i need help with this :sad::sad:. I have this function u(w) = 10w - 0,001w2

And i need to show that for the stochastic variable that:

E [u(w)] = 10µw − 0.001µw2 - 0.001σw2

µw = E [W]
σw2 = var(W)

It's the mean of a discrete random variable W, I'm not sure how they get to that result :(.
 
Last edited:
It looks straight forward to me. I presume you know that (assuming w is a continuous variable) that \(\displaystyle E(f(w))= \int f(w)p(w)dw\), where p(w) is the pdf for w, and \(\displaystyle \sigma^2(w)= \int (w- E(w))^2p(w)dw\). (If w is discrete those integrals are sums but we can use the same notation.)

So \(\displaystyle E(u(w))= \int (10w+ 0.001w^2)p(w)dw= 10\int w p(w)dw+ 0.001\int w^2p(w)dw\)

Now do that integration.
 
HallsofIvy said:
It looks straight forward to me. I presume you know that (assuming w is a continuous variable) that \(\displaystyle E(f(w))= \int f(w)p(w)dw\), where p(w) is the pdf for w, and \(\displaystyle \sigma^2(w)= \int (w- E(w))^2p(w)dw\). (If w is discrete those integrals are sums but we can use the same notation.)

So \(\displaystyle E(u(w))= \int (10w+ 0.001w^2)p(w)dw= 10\int w p(w)dw+ 0.001\int w^2p(w)dw\)

Now do that integration.
Click to expand...

Oh sorry i get E [u(w)] = 10µw + 0.001µw​2, sorry do you have time to explain the steps, sorry
 
Last edited:
You must log in or register to reply here.
Top