uniroot() problem

Thanks for this reply. Here I was trying to calculate implied volatility
using BS formula. This is my code :

oo = 384.40 # traded option price
    uu = 1563.25 # underlying price
    tt = 0.656 # time to maturity in year
    ii = 2.309/100 # interest rate, annualized
    th.price = function(x)
       {
        d1 = (log(uu/K) + (ii + x^2/2)*tt) / (x*sqrt(tt)); d2 = d1 -
x*sqrt(tt)
        option.price = uu * pnorm(d1) - K * exp(-ii*tt) * pnorm(d2)
        return(option.price - oo)
       }

uniroot(th.price, c(-20, 20), tol=1/10^12)

I got following result :

uniroot(th.price, c(-20, 20), tol=1/10^12)

$root
[1] 6.331672e-13

$f.root
[1] 36.28816

$iter
[1] 55

$estim.prec
[1] 7.385592e-13

Hence using implied volatility, difference between traded price and
theoretical price is coming as high as 36.28816, even I increse the
precision level. Any idea how to crack this problem?



Albyn Jones wrote:

One can't tell for sure without seeing the function, but I'd guess  
that you have   a numerical issue.  Here is an example to reflect upon:

f=function(x) (exp(x)-exp(50))*(exp(x)+exp(50))
uniroot(f,c(0,100))

$root
[1] 49.99997

$f.root
[1] -1.640646e+39

$iter
[1] 4

$estim.prec
[1] 6.103516e-05

.Machine$double.eps^0.25/2

[1] 6.103516e-05

uniroot thinks it has converged, at least in relative terms.  Note  
that the estimated precision is related to the machine epsilon, used  
in the default value for "tol".  try fiddling with the tol argument.

uniroot(f,c(0,100),tol=1/10^12)

$root
[1] 50

$f.root
[1] 1.337393e+31

$iter
[1] 4

$estim.prec
[1] 5.186962e-13

albyn


Quoting megh <megh700004 at yahoo.com>:

I have a strange problem with uniroot() function. Here is the result :

uniroot(th, c(-20, 20))

$root
[1] 4.216521e-05

$f.root
[1] 16.66423

$iter
[1] 27

$estim.prec
[1] 6.103516e-05

Pls forgive for not reproducing whole code, here my question is how
"f.root"
can be 16.66423? As it is finding root of a function, it must be near
Zero.
Am I missing something?

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