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Given $f(x), g(x): \mathbb{R} \to \mathbb{R}$ we can form the convolution $f * g$. Define $h(x) = f(cx)$ for some $c>0$. Can we express the convolution $h*g$ in terms of $f * g$ ? Thanks for all help... I am trying to use this in independent study of probability.

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up vote 1 down vote accepted

No but what you get is this: $$h*g(x) = \int h(y)g(x-y)dy = \int f(cy)g(x-y)dy.$$ Substitute $z=cy$ and get $$h*g(x) = \frac{1}{c}\int f(z)g((cx - z)/c)dz.$$

In other words: With $\phi(x) = g(x/c)$ it holds $$h*g(x) =\frac{1}{c} f*\phi(cx).$$

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