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My question is in the context of probability distributions, whose Fourier transforms (characteristic function) almost always exit.

If $f(x)$ be some function such that $ \int_{-\infty}^\infty f(x) \, dx=1,$

then what can be said/deduce about $ \frac{1}{a} \int_{-\infty}^\infty f\left(\frac{1}{a}x\right) \, dx,$ where $a \in \mathbb{R^+}?$

Michael Hardy
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kaka
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1 Answers1

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Hint: use change of variable $z=\frac{x}{a}$.

Kim Jong Un
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  • It turns out that $ \frac{1}{a} \int_{-\infty}^{\infty} f(\frac{1}{a}x)dx=1$ too. Hence it is also a probability distribution. IS this correct? – kaka Oct 13 '14 at 01:21
  • $\frac{1}{a}f(\frac{1}{a}x)$ can indeed be considered a probability density (provided that $f\geq 0$). – Kim Jong Un Oct 13 '14 at 01:22
  • yes ofcourse, if $f(x)$ is some distribution then it is for sure that $f \ge 0$ and so is $f(ax) \ge 0.$ – kaka Oct 13 '14 at 01:26