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I have some problems to show the following relation, apparently using integration by parts and knowing that $\phi$ denotes the density of the standard one dimensional normal distribution. $$\int g(\omega) \delta_x (\omega) \phi\left(\frac{\omega}{\sqrt{t}}\right)~d\omega = \int g(\omega) \mathbf 1_{\left[x,\infty\right]} (\omega) \frac{\omega}{t}\phi\left(\frac{\omega}{\sqrt{t}}\right)~d\omega- \int g'(\omega) \mathbf 1_{\left[x,\infty\right]} (\omega) \phi\left(\frac{\omega}{\sqrt{t}}\right)~d\omega$$

Could someone enlight me please?

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1 Answer 1

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With $f(\omega)=\delta_x(\omega)$ and $h(\omega)=\phi(\omega/\sqrt t)$, we have $h'(\omega)=-(\omega/t)\phi(\omega/\sqrt t)$ and $F(\omega)=\mathbf1_{[x,\infty]}$ with $F'(\omega)=f(\omega)$, and your equation becomes

$$ \int g(\omega)f(\omega)h(\omega)\mathrm d\omega=-\int g(\omega)F(\omega)h'(\omega)\mathrm d\omega-\int g'(\omega)F(\omega)h(\omega)\mathrm d\omega $$

and thus

$$ \int f(\omega)\left(g(\omega)h(\omega)\right)\mathrm d\omega=-\int F(\omega)(g(\omega)h(\omega))'\;. $$

There's no boundary term because $\delta$ vanishes at the boundaries.

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