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I have some questions regarding a proof of Gÿongy's lemma given in 1

I would like to understand the following passage: $$ \int_{s=t_0}^{s=t}\mathbb{E}\left[\delta(X_s-K)\langle dX_s\rangle^2 \right]= \int_{s=t_0}^{s=t}\mathbb{E}\left[\delta(X_s-K) \right] \mathbb{E}\left[\langle dX_s\rangle^2|X_s=K \right] $$

Thanks

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If $X_s \neq K $ then the delta function gives zero, and the product is zero. So the term only contributes when $X_s=K$.

Re-comment, the key to understanding this is the conditional expectation:

$ E \left[ dX_s^2 \mid X_s=K \right] =\frac{E\left[ dX_s^2 \delta(X_s-K)\right]}{E\left [\delta(X_s-K)\right] }$

Where it might be helpful if you interpret the delta as indicator of $X_s=K$

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  • $\begingroup$ What I don't understand if why the expectation of a product equals the product of expectations. Is this true because the quadratic variation $\langle dX_s\rangle^2$ is independant from $X_s$ ? $\endgroup$
    – Aguelmame
    Apr 22 '19 at 17:04
  • $\begingroup$ Added further explanation in the answer. $\endgroup$ Apr 22 '19 at 17:29
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    $\begingroup$ Ok, that makes sense. Thanks $\endgroup$
    – Aguelmame
    Apr 22 '19 at 18:09

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