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The convolution of $f$ and $g$ is defined as $ (f * g )(t) \, \stackrel{\mathrm{def}}{=}\ \int_{-\infty}^\infty f(\tau)\, g(t - \tau) \, d\tau $.

Let's say that $f(t)$ and $g(t)$ have units of, say, meters and Hertz, respectively. What units does the convolution $(f * g )(t)$ have?

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    $\begingroup$ Well a integral behaves like a sum. So if $t$ and $\tau$ are time in seconds the result is in meters * Hertz * seconds = meters. Then $g$ is a transfer function transforming a signal expressed in meters. $\endgroup$ – Yves Mar 20 '17 at 7:38
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    $\begingroup$ For this question to be answerable, you also need to specify the units of $\tau$. Regardless, this question seems to have little to do with statistics: whenever such convolutions arise in statistics, $f$ and $g$ are either probabilities or probability densities, not physical quantities. Perhaps you would get more relevant answers by migrating your question to Physics? $\endgroup$ – whuber Mar 20 '17 at 13:40
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    $\begingroup$ @whuber A probability density has a unit depending on the r.v. of interest as do moments, regression coefficients and most of concepts used by statisticians. These units are not necessarily from physics and can relate e.g. to economics. While most textbooks of applied statistics do not care much about this, it may be worth a discussion. Maybe a question on this topic could help? $\endgroup$ – Yves Mar 25 '17 at 9:56
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    $\begingroup$ @Yves I think you're right. I have answered several questions that needed only a basic understanding of units to resolve and have been thinking that a canonical thread that lays out the "units calculus" could be useful. $\endgroup$ – whuber Mar 25 '17 at 14:23
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Let $X$ and $Y$ be independent random variables with densities $f, g$ respectively. Then the convolution of $f$ and $g$ $$ (f * g )(t) \, \stackrel{\mathrm{def}}{=}\ \int_{-\infty}^\infty f(\tau)\, g(t - \tau) \, d\tau $$ is the density function of $X+Y$. For this to make dimensional meaning, we must assume that $X$ and $Y$ are measured with the same units of measurement, it could for instance be m/s (meters per second). Then the unit of measurement of the probability density function is probability per (m/s). Since probability is a pure number we can write this as 1/(m/s). Let us make this more general by writing u for whatever common unit of measurement of $X$ and $Y$, then the densities $f,g$ has units 1/u.

We can represent the convolution integral above with a Riemann sum: $$ = \sum_i f(\tau_i) g(t-\tau_i) (\tau_i -\tau_{i-1}) $$ The variable $\tau$ above clearly has unit u. So the unit of measurement of each term in the Riemann sum has unit $$ \frac{1}{\text{u}}\cdot \frac{1}{\text{u}}\cdot\text{u} = \frac{1}{\text{u}} $$ showing that the density function obtained by convolving $f$ and $g$ has the same unit as $f$ and $g$.

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