I want to differentiate w.r.t. $\sigma^2$ the following equation
$u'(Y)\mu$ + $\frac{u''(Y)}{2}$$(\sigma^2 + \mu^2) = 0$
where we can consider $\mu$(reward) as an implicit function of $\sigma^2$(risk) of small bets
u'(Y)$\frac{du}{d\sigma^2} +\frac{u''(Y)}{2} + \mu u''(Y) \frac{du}{d\sigma^2}=0$
$u'(\sigma^2)=\frac{u''(Y)}{2(u'(Y) + \mu u''(Y)} $
But answer given in the PDF downloaded from the internet is
$u'(\sigma^2)=\frac{u''(Y)}{2(u'(Y) + \mu(\sigma^2) u''(Y)}$
Which answer is correct?
My another question is what does 1st derivative and 2nd derivative of Utility function imply?

My second question is as follows:

Consider a financial market with one single period, with interest rate r and one stock S. Suppose that $S_0 =1$ and, for n=1, $S_1$ can take two different values: 2, 1/2.

For which values of r the market is viable? viable means (free of arbitrage opportunities).

What if $S_1$ can also take the value 1.

Solution provided.

We want to calculate the values of r such that there is an arbitrage opportunity. We take a portfolio with zero initial value $V_0=0$ Then we invest the amount q in the stock without risk, we have to invest −q in the risky stock (q can be negative or positive). We calculate the value of this portfolio in time 2.



So, if r > 1 there is an arbitrage oppportunity taking q positive (money in the bank account and short position in the risky stock) and if r < −1/2 we have an arbitrage opportunity with q positive (borrowing money and investing in the risky stock). The situation does not change if $S_1$ take the value 1.

I want to know,Now if r<-1/2 how there is an arbitrage opportunity?Would any member explain me in his reply?

That means to have a viable market, $r$ must be between -0.5 To 1

  • $\begingroup$ Please don't ask two questions in one but post separate questions. $\endgroup$ – LocalVolatility Mar 8 at 11:12

$$ \mu = u(\sigma^2) $$ diffentiating $$ diff(u(\sigma^2)^2) = 2u(\sigma^2)u'(\sigma^2) =2\mu(\sigma^2)u'(\sigma^2) $$

both answer are right


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