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I want to calculate $E_t[(X_T-K)^+]$ where $$dX_t=\frac{3}{X_t}dt+2X_t dW_t$$ and $X_0=x$. I don't know how extact the strong solution of this SDE. Indeed I used Ito's lemma but it was not usefule.

Thanks for attantion.

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2 Answers 2

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Note that \begin{align*} d\left(X_t^2\right) &= 2X_t dX_t + d\langle X, X\rangle_t\\ &=(6+4X_t^2)dt + 4X_t^2dW_t, \end{align*} which can be solved using the technique with an integral factor. Specifically, note that \begin{align*} d\left(e^{4t-4W_t}X_t^2 \right) &= X_t^2 d\left(e^{4t-4W_t}\right) + e^{4t-4W_t} d(X_t^2) + \left\langle d\left(e^{4t-4W_t}\right), d(X_t^2) \right\rangle\\ &=e^{4t-4W_t}X_t^2(12dt-4dW_t)\\ &\quad +e^{4t-4W_t}\left[(6+4X_t^2)dt + 4X_t^2dW_t\right]- 16 e^{4t-4W_t}X_t^2 dt\\ &=6e^{4t-4W_t} dt. \end{align*} Then \begin{align*} e^{4t-4W_t}X_t^2 &= x^2+6\int_0^t e^{4s-4W_s} ds. \end{align*} That is, \begin{align*} X_t^2 = e^{4W_t-4t}\left(x^2+6\int_0^t e^{4s-4W_s} ds \right). \end{align*}

A general note. For an equation of the form \begin{align*} dY_t = (aY_t+b)dt + (cY_t+d) dW_t, \end{align*} we can apply an integral factor of the form $$e^{(-a+\frac{1}{2}c^2)t -cW_t}.$$

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  • $\begingroup$ Thanks. Why should we use dynamic of $X_t^2$ instead of dynamic of $X_t^3$ or $X_t^4$? $\endgroup$
    – user22715
    Commented Nov 2, 2016 at 18:43
  • $\begingroup$ That is based on the structure of your equation. With $X_t^2$, you can have an equation that have coefficients as linear functions of $X_t^2$. For $X_t^3$ and $X_t^4$, they do not have such nice property. The equation for $X_t^2$ does not come up arbitrarily: if you multiply $X_t$ to both sides of your equation, you instantly see the relationship with $X_t^2$. $\endgroup$
    – Gordon
    Commented Nov 2, 2016 at 18:49
  • $\begingroup$ How should we know it? $\endgroup$
    – user22715
    Commented Nov 2, 2016 at 18:57
  • $\begingroup$ This equation $X_tdX_t = 3dt +2 X_t^2 dW_t$ is the same as your original equation. Do you see the relationship with $X_t^2$? You may need a relatively good sense in mathematics. $\endgroup$
    – Gordon
    Commented Nov 2, 2016 at 19:01
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You shouldn't say: "Ito's lemma wasn't useful". Set $$Y_t=-2W_t+2t\tag 1$$ Note $W_0=0$ thus $Y_0=0$.We have $$dY_t=2\,dt-2\,dW_t\tag 2$$ Set $Z_t=e^{Y_t}$. By application of Ito's lemma, we have $$dZ_t=e^{Y_t}\,dY_t+\frac{1}{2}e^{Y_t}d[Y_t,Y_t]\tag 3$$ therefore $$dZ_t=4e^{Y_t}dt-2\,e^{Y_t}dW_t=4Z_tdt-2Z_tdW_t\tag 4$$ on the other hand $$d(X_t\,Z_t)=Z_t\,dX_t+X_t\,dZ_t+d[X_t,Z_t]\tag 5$$ thus $$d(X_t\,Z_t)=\frac{3Z_t}{X_t}dt\tag 6$$ in other words $$d(X_t\,Z_t)=\frac{3Z_t^2}{X_tZ_t}dt\tag 7$$ set $R_t=X_tZ_t$. We have $$dR_t=\frac{3Z_t^2}{R_t}dt\tag 8$$ This can be solved as a regular ODE with separable variables : $$R_t\,dR_t=3Z_t^2\, dt$$ and $$\frac{1}{2}R_t^2-\frac{1}{2}R_0^2=3\int_0^{t}Z_s^2ds\\ \frac{1}{2}R_t^2-\frac{1}{2}x^2=3\int_0^{t}Z_s^2ds\tag 9$$ then $$X_t^2=e^{-2Y_t}\left(x^2+6\int_0^{t}e^{2Y_s}ds\right)\tag {10}$$ Finally

$$\color{red}{X_t^2=e^{4W_t-4t}\left(x^2+6\int_0^{t}e^{-4W_s+4s}ds\right)\tag{11}}$$

Now How can you calculate $\mathbb{E}_t[(X_T-K)^+]?$

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  • $\begingroup$ Sorry Gordon's solution is easier than yours. $\endgroup$
    – user22715
    Commented Nov 2, 2016 at 20:04
  • $\begingroup$ No problem --:) $\endgroup$
    – user16651
    Commented Nov 2, 2016 at 20:06

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