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In the Black Scholes framework what is the value of an at-the-money vanilla European call option as time to maturity goes to infinit ($T \rightarrow \infty$)?

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    $\begingroup$ You should at the very least show what you have tried so far and where you are stuck. We are not here to do you exercise sets. $\endgroup$ – SRKX Nov 7 '16 at 6:39
  • $\begingroup$ @SRKK, I am sorry if it looks like i am asking you guys for solving my questions. I didnot get any ideas how to approach this one but when Gordon posted the answer i posted my thinking as well. I will keep this in min $\endgroup$ – Raveesh Nov 7 '16 at 14:46
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The Black-Scholes call option price is given by \begin{align*} C = S_0 N(d_+) - K e^{-rT}N(d_-), \end{align*} where $$d_{\pm}= \frac{\ln \frac{S_0}{K}+(r\pm\frac{1}{2}\sigma^2) T}{\sigma \sqrt{T}}.$$ Here, we assume that the interest rate $r\ge 0$ and $\sigma >0$. For an at-the-money call option, that is, $K=S_0$, we note that $\lim_{T\rightarrow \infty} d_+ = \infty$, that is, $$\lim_{T\rightarrow \infty} N(d_+) = 1.$$ In addition, if $r>0$, then $$\lim_{T\rightarrow \infty}Ke^{-rT} N(d_-) = 0,$$ and if $r =0$, then $\lim_{T\rightarrow \infty} d_- = -\infty$, and $\lim_{T\rightarrow \infty} N(d_-) = 0.$ Therefore, $$\lim_{T\rightarrow \infty} C = S_0.$$

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    $\begingroup$ only if r is less than one half sigma squared.... $\endgroup$ – user9403 Nov 6 '16 at 18:14
  • $\begingroup$ Thanks so much, If q>0, then would the value of the call option be 0? $\endgroup$ – Raveesh Nov 6 '16 at 18:45
  • $\begingroup$ Pardon my ignorance but wouldnt $\lim_{T\rightarrow \infty} e^{-qt} = 0$ and thus the value of the call option be zero $\endgroup$ – Raveesh Nov 6 '16 at 18:59
  • $\begingroup$ You may need to show your work. What is $q$ and how does it appear in Black-Scholes? $\endgroup$ – Gordon Nov 6 '16 at 20:43

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