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Given a random variable Y, is it possible to construct a martingale M such that $$M_1 \stackrel{D}{=} Y$$

I'm not sure how to go about proving that such an M exists under such general conditions, but I feel like this is definitely a standard question...

john
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1 Answers1

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If random variable $Y$ takes values in $\mathbb R$, you can find a measurable function $f: \mathbb R\to \mathbb R$, such that $f(N)$ has the same distribution as $Y$, where $N$ is a standard Gaussian random variable.

Now, for a Brownian motion $B_t$, define $$ M_t = E(f(B_1)|\mathcal F^B_t)$$

Jay.H
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  • How is that first result proved? The one where you can find a measurable $f$ such that $f(N)$ has the same distribution as $Y$? – john Apr 09 '16 at 00:05
  • Let $g$ be the cdf of $Y$, ang $h$ that of N, and $f= g^{-1}h$ – Jay.H Apr 09 '16 at 00:19