Am I misinterpreting this example of convolution?

Question

I am reading the MIT Deep Learning textbook's description of convolutional neural networks and they introduce the mathematical operation of convolution as follows:

Suppose we are tracking the location of a spaceship with a laser sensor. Our laser sensor provides a single output x(t), the position of the spaceship at time t. Both x and t are real valued, that is, we can get a different reading from the laser sensor at any instant in time. Now suppose that our laser sensor is somewhat noisy. To obtain a less noisy estimate of the spaceship’s position, we would like to average several measurements. Of course, more recent measurements are more relevant, so we will want this to be a weighted average that gives more weight to recent measurements. We can do this with a weighting function w(a), where a is the age of a measurement. If we apply such a weighted average operation at every moment, we obtain a new function s providing a smoothed estimate of the position of the spaceship: $$s(t)=\int x(a)w(t-a)da$$

Wouldn't it make way more sense to define the integrand as $x(t-a)w(a)$ in this case, since we want the position function to have some reference of the time value for which we are trying to approximate position? Is this simply a typo in the book, or am I misunderstanding?

Answer 1

The correct way to write the convolution is with a definite integral: $$s(t)=\int_{-\infty}^\infty x(a)w(t-a)da$$ Now let's do a change of variables, such as $b=t-a$. The limits will switch order and $db=-da$ $$s(t)=\int_{\infty}^{-\infty}x(t-b)w(b)(-db)=\int_{-\infty}^{\infty}x(t-b)w(b)db$$ As mentioned in the comment, the convolution is commutative.

Answer 2

As pointed out in the comments and the other answer, the result is numerically identical.

However, you are correct that your version of the integrand makes more sense, given the textual description in the textbook including the introduction of the variable $a$ as the "age" of the observed data, rather than the time of the observed data. If we want to weight each observation by its age $a$, then the integrand should consist of the weight associated with that age $w(a)$ and the observation that has age $a$ at time $t$ which is $x(t-a)$. This gives your version of the integrand $x(t-a)w(a)$

The textbook integral still makes intuitive sense, if you ignore the textual description of the variable $a$ as the "age" of the observation and instead consider it the time of the observation. Then, if we want to weight each observation collected at time $a$ by its age, then the integrand should consist of the observation collected at time $a$ which is $x(a)$ multiplied by the weight associated with its age. Since its age is $t-a$, its weight is $w(t-a)$. This gives the textbook version of the integrand $x(a)w(t-a)$.