Conditional distribution of multivariate normal distribution

Question

I'm doing some self studying this, I got stuck on this question:

Suppose $(X,Y,Z)'$ is normal with density \begin{equation*} \begin{split} C\cdot \text{exp}\{-\frac{1}{2}(4x^{2}+3y^{2}+5z^{2}+2xy+6xz+4zy)\}, \end{split} \end{equation*} where $C$ is a normalizing constant. Determine the conditional distribution of $X$ given that $X+Z=1$ and $Y+Z$=0.

This is what I tried:

\begin{equation*} \begin{split} \mathbf{\mu}=\left[\begin{array}{c} 0 \\ 0\\ 0 \end{array}\right]\qquad \Lambda^{-1}=\left[\begin{array}{ccc} 4& 1& 3 \\ 1& 3& 2\\ 3& 2& 5 \end{array}\right],\qquad \\ \begin{array}{c} U=X \\ V=X+Z \\ W=Y+Z \end{array} \\ \mathbf{B}=\left[\begin{array}{c} U \\ V\\ W \end{array}\right]=\left[\begin{array}{ccc} 1 & 0& 0 \\ 1 & 0& 1\\ 0& 1& 1 \end{array}\right] \end{split} \end{equation*} \begin{equation*} \begin{split} \mathbf{B\mu}=\left[\begin{array}{ccc} 1 & 0& 0 \\ 1 & 0& 1\\ 0& 1& 1 \end{array}\right]\left[\begin{array}{c} 0 \\ 0\\ 0 \end{array}\right] =\left[\begin{array}{c} 0 \\ 0\\ 0 \end{array}\right] \end{split} \end{equation*} \begin{equation*} \begin{split} \Lambda=\frac{1}{|\mathrm{det} \Lambda^{-1}|} \left[\begin{array}{ccc} \mathrm{det} \left|\begin{array}{cc} 3 &2 \\ 2 &5 \end{array}\right|=11 &\mathrm{det}\left|\begin{array}{cc} 1 &2 \\ 3 &5 \end{array}\right|=(-1)& \mathrm{det}\left|\begin{array}{cc} 1 &3 \\ 3 &2 \end{array}\right|=(-7) \\ \mathrm{det}\left|\begin{array}{cc} 1 &3 \\ 2 &5 \end{array}\right|=(-1) & \mathrm{det}\left|\begin{array}{cc} 4 &3 \\ 3 &5 \end{array}\right|=11 &\mathrm{det}\left|\begin{array}{cc} 4 &1 \\ 3 &2 \end{array}\right|=5 \\ \mathrm{det}\left|\begin{array}{cc} 1 &3 \\ 3 &2 \end{array}\right|=(-7) & \mathrm{det}\left|\begin{array}{cc} 4&3 \\ 1 &2 \end{array}\right|=5 & \mathrm{det}\left|\begin{array}{cc} 4 &1 \\ 1 &3 \end{array}\right|=11 \end{array}\right]\\ \frac{1}{|\mathrm{det}\Lambda^{-1}|}=4\cdot 3\cdot 5-1\cdot 1\cdot 5-2\cdot 2\cdot 4-3\cdot 3\cdot 3+1\cdot 2\cdot 3+1\cdot 2\cdot 3\\ =60-5-16-27+6+6=24\\ \Lambda=\frac{1}{24}\left[\begin{array}{ccc} 11 & (-1) & (-7) \\ (-1) & 11 & 5\\ (-7) & 5 & 11 \end{array}\right] \end{split} \end{equation*} \begin{equation*} \begin{split} \mathbf{B \Lambda}=\frac{1}{24}\left[\begin{array}{ccc} 1 & 0& 0 \\ 1 & 0& 1\\ 0& 1& 1 \end{array}\right]\left[\begin{array}{ccc} 11 & (-1) & (-7) \\ (-1) & 11 & 5\\ (-7) & 5 & 11 \end{array}\right]\\ =\frac{1}{24}\left[\begin{array}{ccc} 1\cdot 11+0\cdot (-1)+0\cdot (-7) & 1\cdot (-1)+0\cdot 11+0\cdot 5& 1\cdot (-7)+0\cdot 5+0\cdot 11 \\ 1\cdot 11+0\cdot (-1)+1\cdot (-7) & 1\cdot (-1)+0\cdot 11+1\cdot 5& 1\cdot (-7)+0\cdot 5+1\cdot 11\\ 0\cdot 11+1\cdot (-1)+1\cdot (-7) & 0\cdot (-1)+1\cdot 11+1\cdot 5 &0\cdot (-7)+1\cdot 5+1\cdot 11 \end{array}\right]\\ =\frac{1}{24}\left[\begin{array}{ccc} 11 & (-1) & (-7) \\ 4 & 4 & 4\\ (-8) & 16 & 16 \end{array}\right] \end{split} \end{equation*} \begin{equation*} \begin{split} \mathbf{B\Lambda B^{T}}=\frac{1}{24}\left[\begin{array}{ccc} 11 & (-1) & (-7) \\ 4 & 4 & 4\\ (-8) & 16 & 16 \end{array}\right]\left[\begin{array}{ccc} 1 & 1&0 \\ 0 &0 &1\\ 0& 1&1 \end{array}\right]\\ =\frac{1}{24}\left[\begin{array}{ccc} 11\cdot 1+(-1)\cdot 0+(-7)\cdot 0 & 11\cdot 1+(-1)\cdot 0+(-7)\cdot 1& 11\cdot 0+(-1)\cdot 1+(-7)\cdot 1 \\ 4\cdot 1+4\cdot 0+4\cdot 0 & 4\cdot 1+4\cdot 0+4\cdot 1& 4\cdot 0+4\cdot 1+4\cdot 1\\ (-8)\cdot 1+16\cdot 0+16\cdot 0& (-8)\cdot 1+16\cdot 0+16\cdot 1& (-8)\cdot 0+16\cdot 1+16\cdot 1 \end{array}\right]\\ =\frac{1}{24}\left[\begin{array}{ccc} 11 & 4& (-8) \\ 4 & 8& 8\\ (-8)& 8& 32 \end{array}\right] \end{split} \end{equation*} \begin{equation*} \begin{split} \mathrm{E}(U|V=0,W=1)=\mu_{x}+\Sigma_{12}\Sigma_{22}^{-1}\left[\begin{array}{c} 1-\mu_{v}\\ 0-\mu_{w} \end{array}\right]\\ \Sigma_{12}=\left[\begin{array}{cc} 4&(-8) \end{array}\right]\\ \Sigma_{21}=\left[\begin{array}{c} 4 \\ (-8) \end{array}\right]\\ \Sigma_{22}=\left[\begin{array}{cc} 8 & 8 \\ 8 & 32 \end{array}\right] \end{split} \end{equation*} \begin{equation*} \begin{split} \Sigma_{22}^{-1}=\frac{1}{8\cdot 32-8\cdot 8}\left[\begin{array}{cc} 32 & -8 \\ -8 & 8 \end{array}\right]=\frac{1}{3\cdot 8\cdot 8}\left[\begin{array}{cc} 32 & -8 \\ -8 & 8 \end{array}\right] \end{split} \end{equation*} \begin{equation*} \begin{split} \mathrm{E}(U|V=1,W=0)=0+\frac{1}{3\cdot 8\cdot 8}\left[\begin{array}{cc} 4&(-8) \end{array}\right]\left[\begin{array}{cc} 32 & -8 \\ -8 & 8 \end{array}\right]\left[\begin{array}{c} 1-\mu_{v}\\ 0-\mu_{w} \end{array}\right]\\ =\frac{1}{3\cdot 8\cdot 8}\left[\begin{array}{cc} 4\cdot 32+(-8)\cdot (-8) & 4\cdot (-8)+(-8)\cdot 8 \end{array}\right]\left[\begin{array}{c} 1-\mu_{v}\\ 0-\mu_{w} \end{array}\right]\\ =\frac{1}{3\cdot 8\cdot 8}\left[\begin{array}{cc} 192 &(-96) \end{array}\right]\left[\begin{array}{c} 1-0\\ 0-0 \end{array}\right]\\ =\frac{1}{192}\left[\begin{array}{cc} 192\cdot 1 + (-96)\cdot 0 \end{array}\right]=1 \end{split} \end{equation*} \begin{equation*} \begin{split} 24\cdot \mathrm{Var}(U|V=1,W=0)=\Sigma_{11}-\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}=11-\frac{1}{192}\left[\begin{array}{cc} 192 + (-96) \end{array}\right]\left[\begin{array}{c} 4 \\ (-8) \end{array}\right]\\ =11-\frac{1}{192}\left[\begin{array}{c} 192\cdot 4+(-96)\cdot (-8) \end{array}\right]\\ =11-\frac{1}{192}\left[\begin{array}{c} 1536 \end{array}\right]=11-8=3\\ \mathrm{Var}(U|V=1,W=0)=\frac{3}{24}=\frac{1}{8} \end{split} \end{equation*} \begin{equation*} \begin{split} X|X+Y=1,Y+Z=0\in N(1,\frac{1}{8}) \end{split} \end{equation*}

Is this correct? My book gives me a different answer.

Answer 1

It looks like you're on the right track, but you're working way too hard on this one. Sooner or later algebraic and numerical mistakes will creep into even the best calculations. A good strategy is to minimize the amount calculation: the Principle of Mathematical Laziness. A key element of this principle is just-in-time computation: don't do any work until you have to. The following solution illustrates these ideas.

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You have seen the virtue of changing variables. Keeping $X$ (whose conditional distribution we wish to compute), let the two new variables be

$$U = Y+Z,\ V = X+Z.$$

Consequently, looking ahead to the next step, note that the original variables can be expressed as

$$Y = U-V+X,\ Z=V-X.$$

You also recognized the need to compute the Jacobian of this transformation. Using the method I have described at https://stats.stackexchange.com/a/154298/919 this is almost trivial:

$$\left|\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z\right| = \left|\mathrm{d}x\wedge \mathrm{d}(u-v+x)\wedge \mathrm{d}(v-x)\right| = \left|\mathrm{d}x\, \mathrm{d}u\, \mathrm{d}v\right|.$$

This leaves only the argument of the exponential, into which we need to substitute

$$y = u-v+x,\ z = v-x$$

and then set $u=0$ and $v=1.$ Focusing on the argument of the exponential (and ignoring the necessary division by $2$), this can be performed by visual inspection of the coefficients of $x$ and $x^2$ and then, as always with Normal distributions, completing the square:

$$\begin{aligned} 4x^{2}&+3y^{2}+5z^{2}+2xy+6xz+4zy\\ &= 4x^2 + 3(u-v+x)^2 + \cdots + 4(v-x)(u-v+x)\\ &= (4+3+5+2-6-4)x^2 \\&+ (0-6-10-2+6+8)x \\&+ \text{constants}\\ &= 4x^2 - 4x + \text{constants} \\ &= \frac{(x-1/2)^2}{(1/2)^2}+\text{some constant}. \end{aligned}$$

We know the conditional distribution will be Normal with some mean $\mu$ and some standard deviation $\sigma$, which means this quadratic part will take the form $(x-\mu)^2/\sigma^2$ plus some constant. Comparing with the foregoing, you can read off the values $\mu=1/2$ and $\sigma=1/2.$

You will, of course, wish to check this work: but I hope you find this to be much less effort than checking your original calculations.