Calculate $\int_{\mathbb{R}^2} uvf(u,v)dudv$

For my probability course I have to work out the following integral and I want to be sure that I did it properly since my last calculus-related course was at least 2 years ago.

The integral is this one:

$$I = \int_{\mathbb{R}^2} uvf(u,v)\mathrm{d}u\mathrm{d}v$$

$$f(u,v) = \frac{1}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}g(u,v)$$

$$g(u,v) = \exp\left( -\frac{1}{2} \frac{1}{1-\rho^2}\left\{\left(\frac{u}{\sigma_1}\right)^2 -2\rho\left(\frac{u}{\sigma_1}\right)\left(\frac{v}{\sigma_2}\right)+\left(\frac{v}{\sigma_2}\right)^2\right\}\right)$$

First, if I define: $U = \frac{u}{\sigma_1}$ amd $V = \frac{v}{\sigma_2}$ I can rewrite the integral the following way:

$$I = (\sigma_1\sigma_2)^2\int_{\mathbb{R}^2}UVf(U\sigma_1,V\sigma_2)\mathrm{d}U\mathrm{d}V$$

The current problem is that we can't separate the two integrals because of the term $UV$ in $g(U\sigma_1,V\sigma_2).$ So, that means that I have to find a change of variables so that I can separate $U$ and $V$.

For that purpose, I want to find a base in which $h(U,V) = U^2 - 2\rho UV + V^2$ is free from the term $UV$.

$$h(U,V) = \pmatrix{U & V} \pmatrix{1 & -\rho \\ -\rho & 1} \pmatrix{U \\ V}$$

Its eigenvalues are: $$\lambda_1 = 1+\rho \text{ and } \lambda_2 = {1-\rho}$$ and its eigenvectors are: $$\vec{\lambda_1} = \pmatrix{1 \\ 1} \text{ and } \vec{\lambda_2} = \pmatrix{1 \\ -1}$$

From there I understand that the variable change $X = U + V$ and $Y = U - V$ will do what I want. Indeed,

$$h(X+Y, X-Y) = 2(X^2(1-\rho) + Y^2(1+\rho))$$

Since $U$ and $V$ ranged over $\mathbb{R}$ it will be the same for $X+Y$ and $X - Y$.

I switch back to lowercase. The integral can now be written:

$$J = \frac{2\pi\sqrt{1-\rho^2}}{\sigma_1\sigma_2}I = \int_{\mathbb{R}^2}(x^2 - y^2)\exp\left\{- \frac{x^2}{1+\rho} - \frac{y^2}{1-\rho} \right\}\mathrm{d}x\mathrm{d}y$$

I can then write,

$$J = J1 - J2$$

$$J1 = \int_{\mathbb{R}^2} x^2\exp\left\{- \frac{x^2}{1+\rho} - \frac{y^2}{1-\rho} \right\}\mathrm{d}x\mathrm{d}y$$

$$J2 = \int_{\mathbb{R}^2} y^2\exp\left\{- \frac{x^2}{1+\rho} - \frac{y^2}{1-\rho} \right\}\mathrm{d}x\mathrm{d}y$$

Before working out $J1$ and $J2$ I remind that:

$$\int_{\mathbb{R}^2} \exp(-ax^2)\mathrm{d}x = \sqrt{\frac{\pi}{a}}$$

and

$$\int_{\mathbb{R}^2} x^2\exp(-ax^2)\mathrm{d}x = \sqrt{\frac{\pi}{4a^3}}$$

First let's work out $J1$.

\begin{align*} J1 &= \int_{\mathbb{R}^2} x^2\exp\left\{- \frac{x^2}{1+\rho} - \frac{y^2}{1-\rho} \right\}\mathrm{d}x\mathrm{d}y \\ &= \int_{\mathbb{R}} x^2\exp\left\{- \frac{x^2}{1+\rho}\right\}\mathrm{d}x\int_{\mathbb{R}} \exp\left\{-\frac{y^2}{1-\rho} \right\}\mathrm{d}y \\ &= \frac{\sqrt{(1+\rho)^3\pi}\sqrt{(1-\rho)\pi}}{2} \end{align*}

Then, I obtain:

$$J2 = \frac{\sqrt{(1-\rho)^3\pi}\sqrt{(1+\rho)\pi}}{2}$$

Thus,

$$J = \frac{\pi (1-\rho^2) (\sqrt{1-\rho} + \sqrt{1+\rho})}{2}$$

Finally, $$I = \frac{\sigma_1\sigma_2}{4} \left( (1-\rho)\sqrt{1+\rho} + (1+\rho)\sqrt{1-\rho} \right)$$

Is this calculation right? I found that question really mean for a starter in a probability class where the lecturer do not know anything about the background of the students. Maybe there is a shorter and simplier solution but I did not saw it.

• I think that the error is when I do the change of variables. I do not multiply by the determinant. – 永劫回帰 Apr 20 '16 at 7:28
• You need the Jacobian of the transformation. – RRL Apr 20 '16 at 7:52
• @RRL: Since my transformation is $U = X + y$ and $V = X - Y$ the Jacobian matrix should be $$\pmatrix{1+y & 1+x \\ 1- y & x - 1}$$ and the Jacobian should be $-2-2xy$. Could you confirm? I would prefer not doing the calculation a second time wrong. – 永劫回帰 Apr 20 '16 at 7:58
• You can't have $U = X + y$ as you are mixing the wrong variables. It has to be $U = X + Y$ and the jacobian is $[1 \,\, 1 ; 1 \,\, -1]$. – RRL Apr 20 '16 at 8:04
• @RRL: Sure that helped me. You showed me how to generalize the problem and how compact notation can be used. Moreover, that's when reading your answer that I found that I forgot to to multiply by the Jacobian and thus I was able to see where my mistake was. || Yes, I mixed uppercase and lowecase in my comment that is an error all should have been uppercase. || Nevertheless since the Jacobian is just a constant with no $\rho$ in it, it means that I have an other error. – 永劫回帰 Apr 20 '16 at 8:09

Using $| \cdot|$ to denote the determinant we have

$$f(u,v) = \frac{|\Sigma|^{-1/2}}{2 \pi}\exp\left(-\frac{1}{2}\mathbb{x}'\Sigma^{-1}\mathbb{x}\right),$$

where

$$\mathbb{x} =\pmatrix{u \\ v},$$

and

$$\Sigma = \pmatrix{\sigma_1^2 & \rho\sigma_1 \sigma_2 \\ \rho \sigma_1 \sigma_2 & \sigma_2^2} \\ \Sigma^{-1} = \frac{1}{1 - \rho^2}\pmatrix{\sigma_1^{-2} & -\rho\sigma_1^{-1} \sigma_2^{-1} \\ -\rho \sigma_1^{-1} \sigma_2^{-1} & \sigma_2^{-2}}.$$

Since $\Sigma^{-1}$ is a positive definite matrix, there is a lower triangular, nonsingular matrix $C$ such that

$$C'\Sigma^{-1}C = I \\ |C'||\Sigma^{-1}||C| = 1 \implies |C| = |\Sigma|^{1/2} \\ I = I^{-1} = (C'\Sigma^{-1}C)^{-1} = C^{-1} \Sigma (C')^{-1} \implies CC' = \Sigma.$$

The actual form is

$$C = \pmatrix{ \sigma_1 & 0 \\ \rho \sigma_2 & \sigma_2\sqrt{1-\rho^2}} .$$

You are trying to find the off-diagonal component of

$$K =\frac{|\Sigma|^{-1/2}}{2 \pi}\int \mathbb{x} \mathbb{x}'\exp\left(-\frac{1}{2}\mathbb{x}'\Sigma^{-1}\mathbb{x}\right) \, d \mathbb{x}.$$

Make the change of variables $\mathbb{x} = C\mathbb{z}.$ The integral transforms to

$$K = \frac{|\Sigma|^{-1/2}}{2 \pi}C\int \mathbb{z}' \mathbb{z}\exp\left(-\frac{1}{2}\mathbb{z}'\mathbb{z}\right) \, |C|d \mathbb{z}C' = C\frac{1}{2 \pi}\int \mathbb{z}' \mathbb{z}\exp\left(-\frac{1}{2}\mathbb{z}'\mathbb{z}\right) \, d \mathbb{z}C'.$$

The double integral factors into single integrals, which are easily computed as

$$\frac{1}{2 \pi}\int \mathbb{z}' \mathbb{z}\exp\left(-\frac{1}{2}\mathbb{z}'\mathbb{z}\right) \, d \mathbb{z} = I.$$

Hence,

$$K = CIC' = CC' = \Sigma.$$

The off-diagonal component is $\rho \sigma_1 \sigma_2.$

• This is a more compact matrix formulation of what you are trying to do here. Your steps should replicate this if you are doing it correctly. – RRL Apr 20 '16 at 6:42
• Yes, thank you for your answer. I will try to match your steps to mine. I just don't understand what do you call off-diagonal component. – 永劫回帰 Apr 20 '16 at 6:44
• My integrand is a matrix with where $x'x = [u^2 \,\,uv ; uv \,\,v^2]$. So the integral you are seeking will be the value in the first row and second column of my integrated matrix. – RRL Apr 20 '16 at 6:47
• That's what I see, your calculation is clearly shorter than mine and you don't have to explicitly find the variable change. – 永劫回帰 Apr 20 '16 at 6:53
• What do you mean by the dash in $C'$ it's not the same matrix as C and its not its inverse too. – 永劫回帰 Apr 20 '16 at 6:56

My last calculus course was about 5 or 6 years ago, so I can't verify your work.

However, it appears that $I = E[UV]$, so I believe we can proceed as follows. It appears that $\mu_u = 0$, $\mu_v = 0$, and I will call $\sigma_u = \sigma_1$, $\sigma_2 = \sigma_v$. Hence, we have $$I = E[UV] = \text{Cov}(U,V)+E[U]E[V] = \text{Corr}(U,V)\text{SD}(U)\text{SD}(V)+0\cdot0 = \rho\sigma_1\sigma_2$$

• I did not considered using the fact that $I = E[UV]$, it appears to be significantly shorter and way less complex. But since we don't have the same results one of us has done a miscalculation (possibly me since my calculation involves way more steps). +1 for suggesting the use of probability tools rather than going head down into the integral. – 永劫回帰 Apr 20 '16 at 5:56
• Well, if the problem was phrased just how you posted it, then it was probably a trick. The professor wants to see how is paying attention, and who is asleep. Haha :) – Em. Apr 20 '16 at 5:57
• Most of the people who know bivariate normal and expecation will follow this route immediately and obtain the answer very quickly. However it is good for you to work out those dirty stuff once. – BGM Apr 20 '16 at 6:00
• Maybe yes, the problem was stated "compute the following expectation value = $E[UV]$" so I should I used that as an hint. – 永劫回帰 Apr 20 '16 at 6:04
• @駑馬十駕 Hahahaha, you stopped thinking. You were asleep. Be careful, don't be a robot! – Em. Apr 20 '16 at 6:08