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I need your help with this problem: Suppose $(X, Y)'$ follows a Bivariate Normal Distribution with parameters $μ_1 ,μ_2, σ_1^2, σ_2^2$, and $ρ$. Let $U = X + Y$ and $V = X - Y$. Considering that $X$ and $Y$ are not independent random variables, how will I get the joint distribution of $U$ and $V$. Thanks in advance!

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The vector $(X,Y)$ is gaussian and $(U,V)=(X+Y,X-Y)$ is a linear function of $(X,Y)$ hence $(U,V)$ is gaussian as well. In particular, the distribution of $(U,V)$ is characterized by its mean vector $M$ and covariance matrix $C$. By definition, $$ M=\begin{pmatrix}\mathbb E(U) \\ \mathbb E(V)\end{pmatrix},\qquad C=\begin{pmatrix}\mathrm{var}(U) & \mathrm{cov}(U,V)\\ \mathrm{cov}(U,V)&\mathrm{var}(V) \end{pmatrix}, $$ with

  • $\mathbb E(U)=\mu_X+\mu_Y$, $\mathbb E(V)=\mu_X-\mu_Y$,
  • $\mathrm{var}(U)=\mathrm{var}(X)+\mathrm{var}(Y)+2\mathrm{cov}(X,Y)=\sigma_X^2+\sigma_Y^2+2\varrho$,
  • $\mathrm{var}(V)=\mathrm{var}(X)+\mathrm{var}(Y)-2\mathrm{cov}(X,Y)=\sigma_X^2+\sigma_Y^2-2\varrho$,
  • $\mathrm{cov}(U,V)=\mathrm{var}(X)-\mathrm{var}(Y)=\sigma_X^2-\sigma_Y^2$.

Except when $\varrho^2=\sigma_X^2\sigma_Y^2$, the distribution of $(U,V)$ has a density $f_{U,V}$, defined by $$ f_{U,V}(u,v)=\frac1{2\pi\sqrt{\det C}}\exp\left(-\frac12\left(u-\mathbb E(U),v-\mathbb E(V)\right)^*C^{-1}\left(u-\mathbb E(U),v-\mathbb E(V)\right)\right). $$

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  • $\begingroup$ Is there really no way to express it using the moment-generating function only? We haven't discussed matrices yet. Thanks anyway. $\endgroup$
    – Mach
    Mar 21 '13 at 0:39
  • $\begingroup$ Could you explain express it using the moment-generating function? (What is "it"? Why the MGF, which appears nowhere in the question??) Density of $(X,Y)$ added to the answer. $\endgroup$
    – Did
    Mar 21 '13 at 7:12
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If I'm not mistaken, the general convolution formula yields that the distribution of $U$, $g(u)$, is given by $$g(u) = \int_{-\infty}^\infty \int_{-\infty}^{u-x} \! f(x,y) \, \mathrm{d}y \mathrm{d}x$$ where $f(x,y)$ is the joint pdf of $X$ and $Y$, i.e. your bivariate normal pdf.

Now, since the Normal distribution is stable, you can use this to your advantage and prove that $X+Y$ will be Normal with mean $\mu = \mu_1 + \mu_2$ and variance $\sigma^2 = \sigma_1^2 + \sigma_2^2 + 2 \rho \sigma_1 \sigma_2$. Similarly, get the distribution of $V$.

These two marginals can be "bound" into a joint density by using a copula function.

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  • $\begingroup$ hmmm, but what's needed is the joint of U and V, i.e. the joint of X + Y and X - Y and I'm pretty sure that will not result in a normal distribution. $\endgroup$
    – Mach
    Mar 20 '13 at 22:15
  • $\begingroup$ Ah, yes, missed that part about needing the joint. So you have the two marginals but need to create the joint from these two. Then, you could use a copula function to bind the two marginals. See the Wikipedia page on copulas. $\endgroup$
    – baudolino
    Mar 20 '13 at 22:42
  • $\begingroup$ Okay, thanks! I assume there's really no way to get the joint directly. I've tried using the moment generating function but it ends up in something that's hard to simplify. $\endgroup$
    – Mach
    Mar 20 '13 at 22:45

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