Hi,

Just wondering on Reading 54, the Value Added = risk*IR - lambda*risk^2

How do they go from that equation to VA(optimal)=IR^2/(4*lambda)?

(I’m getting a negative value as optimal VA using differentiation)

Any help is appreciated thanks.

Hi,

Just wondering on Reading 54, the Value Added = risk*IR - lambda*risk^2

How do they go from that equation to VA(optimal)=IR^2/(4*lambda)?

(I’m getting a negative value as optimal VA using differentiation)

Any help is appreciated thanks.

You should first calculate **W*** which is the optimal residual risk and then using that calculate the optimal VA

Optimal W is IR/(2*lambda)

VA = IR * **W*** - lambda * **W*** ^2

inputing that into VA = IR * IR/(2*lambda) - lambda * (IR^2/(4*lambda^2))

VA = IR^2 /(2*lambda) - IR^2/4*lambda = IR^2/4*lambda

You should first calculate **W*** which is the optimal residual risk and then using that calculate the optimal VA

Optimal W is IR/(2*lambda)

VA = IR * **W*** - lambda * **W*** ^2

inputing that into VA = IR * IR/(2*lambda) - lambda * (IR^2/(4*lambda^2))

VA = IR^2 /(2*lambda) - IR^2/4*lambda = IR^2/4*lambda

The optimal W is basically the first differential (which gives the maxima) of the function. So if you difference w.r.t ‘w’ you get the max VA

I’ve gotten to the second last step but I can’t get to the final step. This is as far as I can get:

IR^2/(2*lambda) - IR^2/(4*lambda) = [IR^2/(2*lambda)][1-(1/(2*lambda))]

IR^2/(2*lambda) - IR^2/(4*lambda) = 2 * IR^2/(4*lambda) - 1 * IR^2/(4*lambda) = ( 2 - 1 ) * IR^2/(4*lambda) = IR^2/(4*lambda)

Thanks MFEMSF, got it now!

You’re welcome!