1. ## inequality

If the expected $/£-rate for a 1 year’s time is 1.50, is the expected rate £/$-rate 2/3, < 2/3 or > 2/3?

(Hint: the expected yearly $/£-rate is defined as the average$/£-daily
rate is: 1/n . sum 1 to n of Xi for i = 1,...n=365 .The expected yearly £/$-rate is 1/n sum 1 to n of 1/Xi WHERE Xi is the$/£-Rate on the ith day.

My first impression is that obviously the expected rate should be the inverse which = 2/3. Im having trouble proving it though:

1/365 ( X1 + X2 + ... + X365) = 3/2

1/365 (1/X1 + 1/X2 + ... + 1/X365) = y

Can equate them by setting 1/365 = 3/2 . 1/(X1 + X2 + ... + X365) = y/(1/X1 + 1/X2 + ... + 1/X365)

But I just can not get anywhere! Any help welcome!

2. I think ive figured it out:

Using Jensens Inequality:

F(E[X]) >= (less than or equal to) E[F(X)]

So, Let F be the function: F(Z) = 1/Z

Then,

F( (1/n) Sum Xi ) >= Sum ( F(X) ) / n
F( (1/n) Sum Xi ) >= Sum ( 1/Xi ) / n
F ( 3/2 ) >= Sum ( F(X) ) / n
2/3 >= Expected value of £/$-Rate Since the function F(Z) is 1-1 then we have equality in the above inequality. Thoughts and opinions welcome! 3. Using Jensen is complete overkill. Try using AM-HM 4. Never heard of this AM-HM, but after some googling - it seems to be exactly Jensens Inequality with F(X)=1/X. Or just some special case of Jensens anyway. 5. Originally Posted by Rog Never heard of this AM-HM, but after some googling - it seems to be exactly Jensens Inequality with F(X)=1/X. Or just some special case of Jensens anyway. Let$\displaystyle a_1,a_2,...,a_n$be positive numbers. Then,$\displaystyle \frac{a_1+...+a_n}{n} \geq \frac{n}{\frac{1}{a_1}+...+\frac{1}{a_n}}\$