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Let $f(x)$ be a continuous probability density function. Show that, for every $-\infty < µ < \infty$ and $σ > 0$, the function $\frac{1}{\sigma}f(\frac{x-\mu}{\sigma})$ is also a probability density function.

Well I know that if $f(x)$ is a cont. probability density fnc, $\int_{-\infty}^{\infty} f(x)dx=1$. Thus I also need to show that $\int_{-\infty}^{\infty}\frac{1}{\sigma}f(\frac{x-\mu}{\sigma})dx$ must be equal to 1.

But in all honesty, I don't know how to do this. I've practiced showing a function to be a c.p.d.f. provided that there is a real function I can work with(not like this abstract one). SO, should I use another definition or some hidden property to get my way around this?

llawliet_78
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2 Answers2

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To qualify as a probability density, the function $h(x):=\frac1\sigma f({x-\mu\over\sigma})$ must be nonnegative and must integrate to $1$. To show that $h$ integrates to one, just do a substitution $t:={x-\mu\over\sigma}$ in the integral $\int_{-\infty}^\infty h(x)\,dx=\int_{-\infty}^\infty \frac1\sigma f({x-\mu\over\sigma})\,dx$.

grand_chat
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I'd write it like this: $$ \int_{-\infty}^\infty f\left( \frac{x-\mu}\sigma \right) \, \left( \frac{dx} \sigma \right) = \int_{-\infty}^\infty f(w)\,dw = \cdots $$

Michael Hardy
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