# Transformation of double-integral with $y-x\leq 1$ and $x-y\leq 1$ for probabilities

Let the the function $f(x,y)$ be given by $$f(x,y)=\begin{cases}cxy,&-1\leq x\leq 0\wedge 0\leq y\leq 1\wedge y-x\leq 1,\\cxy,&0\leq x\leq 1\wedge -1\leq y\leq 0\wedge x-y\leq 1,\\0,&\text{else}.\end{cases}$$ Determine $c$ so that $f(x,y)$ is a probability density function.

My thoughts so far: $$\iint\limits_{-\infty}^\infty f(x,y)\,\mathrm dx\,\mathrm dy=1 \Rightarrow\int\limits_{0}^1\int\limits_{-1}^0cxy\,\mathrm dx\,\mathrm dy+\int\limits_{-1}^0\int\limits_{0}^1cxy\,\mathrm dx\,\mathrm dy=1$$ However the third condition(s) $y-x\leq 1$; $x-y\leq 1$ isn't always satisfied, so i was thinking of a transformation of the integrals by polar coordinate transformations to restrict the integration to the specified area. Do you have any suggestions?

• There are no circles or relatives of circles in the game, so I would not transform to polar. A linear transformation may be useful, or direct integration, paying careful attention to the geometry. – André Nicolas Jun 18 '12 at 15:42
• Draw a picture of the non-zero region to help develop a little intuition, and simply work out the two double integrals. – jbowman Jun 18 '12 at 15:44
• For example, for the first region the integral would become: $\int_{-1}^0(\int_0^{1+x} cxy dy) dx$. – Giovanni De Gaetano Jun 18 '12 at 15:46

Draw a picture. I could not solve the problem without one.

There are two parts to the region where $f(x,y)\ne 0$, the first part in your list and the second part. The two parts are obtained from each other by interchanging the roles of $x$ and $y$, So there will be symmetry across the line $y=x$. The function $cxy$ we are integrating is also symmetric in $x$ and $y$, so the integrals over the two parts will be equal. Therefore we can just integrate over our favourite part, and double the result.

We look in detail at the second part of the region. It turns out that this is the triangle with corners $(0,0)$, $(0,-1)$, and $(1,0)$. At a certain point of the sketching, you will want to know on which side of the line $x-y=1$ our region lies. One way to decide is to rewrite the inequality $x-y \le 1$ as $y \ge x-1$, so $y$ is supposed to be bigger than $x-1$. But $y$ bigger means we are above the line $x-y=1$.

Express our integral over this part as an iterated integral, integrating first with respect to $y$, then with respect to $x$.

The "bottom" curve is $x-y=1$, the top curve is $y=0$. So when we integrate with respect to $y$, we integrate from $y=x-1$ to $y=0$. So we want $$\int_{x=0}^1\left(\int_{y=x-1}^0 cxy\,dy\right)\,dx.$$

The inner antiderivative is $\frac{cxy^2}{2}$. When we substitute our endpoints, we get $-\frac{cx(x-1)^2}{2}$. Now integrate from $x=0$ to $x=1$. To do the integration, you may want to expand out $x(x-1)^2$. And remember that this is half of the ultimate integral. This gives us an excuse to preemptively multiply by $2$.

• Ah, the symmetry is a nice touch. I missed that, since I didn’t have to draw a picture. Shows that pictures can be useful even when they’re not absolutely necessary. – Brian M. Scott Jun 18 '12 at 16:06

As André said, there’s nothing akin to a circle in sight, so polar coordinates aren’t indicated. Besides, it’s easy enough to set up the iterated integrals in rectangular coordinates. Take $$\iint_S f(x,y)\,dA\;,$$ for instance, where $S=\big\{\langle x,y\rangle:0\le x\le 1\text{ and }-1\le y\le 0\big\}$. The only part of $S$ that matters is the part where $f(x,y)\ne 0$, which is where $x-y\le 1$. Equivalently, this is where $y\ge x-1$. Thus, for each $x\in[0,1]$ we’re interested in values of $y$ between $x-1$ and $0$, and we have

$$\iint_S f(x,y)\,dA=c\int_0^1\int_{x-1}^0xy\,dy\,dx\;.$$

This is a straightforward iterated integral. The other double integral can be set up as an equally straightforward iterated integral, and the desired value of $c$ is then easily obtained.

By the way, if I made no mistake in my calculations, $c$ turns out to be negative.

• That $c$ must be negative can be deduced even without setting up the integrals. $f(x,y) = cxy$ in a region where $x$ and $y$ have opposite signs and therefore $c$ must be negative since $f(x,y)$ is a probability density and thus necessarily nonnegative. – Dilip Sarwate Jun 18 '12 at 16:23

Below is the region over which you need to perform your integration.

Note that in the blue region, $x\leq 0$ and $y \geq 0$, whereas in the red region $x \geq 0$ and $y \leq 0$. The integrand is $xy$ and hence the sign of the integrand is negative in both these regions. And hence due to the symmetry, it is enough to perform the integral over one region and double the result to get the answer. Since $f$ is a probability density function, we have that $$\int \int_{\text{Blue region}} cxy dy dx + \int \int_{\text{Red region}} cxy dy dx = 1$$ And by symmetry, we have that $$\int \int_{\text{Blue region}} cxy dy dx = \int \int_{\text{Red region}} cxy dy dx = \dfrac12$$

Let us now perform the integral over the blue region. Let $$I = \displaystyle \int \int_{\text{Blue region}} cxy dy dx$$To perform this integral, we will first fix a $x$ and integrate over $y$ and then integrate over $x$. This is shown in the figure below. For a fixed $x$, as seen from he figure $y$ goes from $0$ to $x+1$. Once we have this, we then vary $x$ from $-1$ to $0$. Hence, \begin{align} I & = \int_{x=-1}^0 \int_{y=0}^{x+1} cxy dy dx = \int_{x=-1}^0 cx \left(\int_{y=0}^{x+1}y dy \right) dx\\& = \int_{x=-1}^0 cx \left. \dfrac{y^2}2 \right \vert_{y=0}^{x+1} dx = \int_{x=-1}^{0} cx \dfrac{(x+1)^2}{2} dx\\& = \dfrac{c}2 \int_{x=-1}^0 (x^3+2x^2 + x) dx = \dfrac{c}2 \left(\left. \dfrac{x^4}4 + 2 \dfrac{x^3}3 + \dfrac{x^2}2 \right)\right \vert_{-1}^0\\& = -\dfrac{c}2 \left( \dfrac14 - \dfrac{2}3 + \dfrac12 \right) = - \dfrac{c}{24}\end{align}Setting this equal to $\dfrac12$, we get that $c = -12$.