Double integration in a region extending below the x-axis

definite integralsintegrationprobability theory

For area under a curve, we separately calculated areas above vs below the x-axis, taking the absolute value for the latter to add to the former.

I'm not sure how this translates to the double integration case. First of all, sorry for asking something can be done by trial but the question I'm working on does not have worked solutions so since I can't get the exact final answer I'm spending hours working blindly.

I have a triangle bounded by $y_1 -1 \leq y_2\leq -y_1+1$ and $0 \leq y_1 \leq 1$. This is where the equation to integrate is $f(y_1, y_2) = 6y_1^2 y_2$ is defined. I'm asked to integrate doubly over $y_1 < \frac{1}{2}, y_2 < \frac{1}{2}$ (this is a probability question where the equation is a pdf and the ask is to find the joint probability of $y_1 < \frac{1}{2}, y_2 < \frac{1}{2}$).

I am guessing the need to take absolute values is different and more complex because although the "area" below x-axis is negative, the z-level is not necessarily negative. Therefore I'm not quite sure how it works.

I've attempted an all-at-once approach which I didn't expect to work and it didn't e.g.
$$
\int_0^{1/2} \int_{y_1 -1}^{1/2} 30y_1^2 y_2 dy_2 dy_1
$$
I've also tried separately doing the parts above and below the x-axis:
$$
\int_{0}^{1/2} \int_{0}^{1/2} 30y_1^2 y_2 dy_2 dy_1 + \int_{0}^{1/2} \int_{y_1 -1}^{0} 30y_1^2 y_2 dy_2 dy_1
$$
But it doesn't seem to be working. Perhaps my algebra is wrong but again with no worked solutions and so far having not found any errors, I have no idea what's going on.

PS the answer is 9/16

Best Answer

I found the answer!!

My method of $$ \int_0^{1/2} \int_{y_1 -1}^{1/2} 30y_1^2 y_2 dy_2 dy_1 $$ was supposed to work, I must have made an algebraic error. Would like to hear an explanation about how $y_2$ cannot be negative though, since I still don't really understand how I got the right answer and what it would mean conceptually.

Related Solutions

[Math] Definite integration and the $y$ axis

If you integrate (I'd rather say, antidifferentiate) $f(x)$, you don't get $g(x)$, you get $g(x)+C$ for some arbitrary constant $C$. So it's wrong to think of $g(b)$ as "the area up to $x=b$."

Let's look at an example. When you antidifferentiate $\sin x$, you could get $-\cos x$, or you could get $-\cos x+42$ (or any of a number of other correct answers, but these two will do for now). So, the area up to $\pi/2$: is it $-\cos(\pi/2)$, which is $0$? or is it $42$?

Well, neither of these is a particularly good answer, which illustrates the difficulties you get into if you think of $g(b)$ as the area up to $x=b$.

Rather, you should think of $\int_a^xf(t)\,dt=g(x)-g(a)$ as the area from $a$ to $x$; it's only definite integrals that have an unambiguous interpretation as areas.

Notice that if we use a new symbol, say, $h(x)=\int_0^xf(t)\,dt$, for the area from zero to $x$ then if $f$ is always positive and $a$ is negative we'll have $h(a)$ negative, so in that sense area to the left of the $y$-axis is negative. But what this really shows is that you can't always interpret $\int_a^bf(x)\,dx$ as an area; you can always calculate the definite integral, but the number that comes out will not be what we generally accept as the area unless $a\le b$ and $f(x)\ge0$ for $a\le x\le b$.

[Math] Gaussian integral with Dirac delta function

I have a Gaussian noise source $V(x,t)=\int d y\, e^{-A (x-y)^2} \xi(x,t)$, where $\xi$ is a Brownian noise, possessing an average of zero $\langle \langle \xi(x,t)\rangle \rangle=0$, and an autocorrelation function $\langle \langle \xi(x_1,t),\xi(x_2,t_2)\rangle \rangle=\gamma\, \delta(x_1-x_2)\delta(t_1-t_2)$.

You must be mistaken, still.

You can rearrange that integral as $V(x, t) = \xi(x, t)\int dy~e^{-A (x - y)^2};$ it cannot fundamentally be a different sort of noise than $\xi$ is, except that it might have a different mean/variance.

Brownian noise does not have a $\delta$ autocorrelation; the (discrete) definition of Brownian noise is (in JavaScript ES6; Python is similar)

function* white_noise() {
    var r, a;
    while (true) { // Box-Muller method
        r = Math.sqrt(-2 * Math.log(Math.random()));
        a = 2 * Math.PI * Math.random();
        yield r * Math.sin(a);
        yield r * Math.cos(a);
    }
}
function* brown_noise(r) { 
    var current = 0, p = Math.sqrt(1 - r*r), white = white_noise();
    while (true) {
        current = r * current + p * white.next().value;
        yield current;
    }
}

and as you can see, it explicitly takes an autocorrelation argument and produces a stream of values which depend on prior history.

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