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By Section 5.5 of the book [Karatzas and Shreve 1991], the following 1-d SDE has unique weak solution in the form of \begin{equation} d X_{t} = X_{t}^{\gamma} \cdot I_{\{X_{t}\ge 0\}} dW_{t}, \ X_{0} =1 \end{equation} for any given constant $\gamma$. It is also well known that if $\gamma = 1$, then $X$ is a true martingale; while if $\gamma = 2$, it is a strict local martingale.

[Q] Is $X$ a true martingale if $\gamma = 5/4$?

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I received recently an friendly invitation to contribute more to the forum. So here are my two cents. Supposing uniqueness in law of equations :

$ dX_{t} = X_{t}^{\gamma} dW_{t}= X_{t}^{\gamma-1}.X_{t} dW_{t}=a(X_t).X_{t} dW_{t}, \ X_{0} =1$ (1)

and

$ dY_{t} = a^{-1}(Y_t).Y_{t}dW_{t}=Y_t^{\frac{1}{\gamma}+1}dW_t, \ Y_{0} =1$ (2)

Then applying theorem 4 (from George Lowther's Blog here), for $\gamma= 5/4$ and as for any $K>0$,
$\int_K^{+\infty}x^{-3/2}dx=[-2\frac{1}{\sqrt{x}}]_K^\infty=\frac{2}{\sqrt{K}}<\infty$

then $X$ is not a martingale.

Best regards

PS: I believe the indicator is not necessary here

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  • $\begingroup$ I put the indicator to avoid ambiguity caused by non-uniqueness. For instance, when $\gamma = 1/4$. $\endgroup$
    – user79963
    Jun 2, 2013 at 0:19

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