Let $Y$ ~ $MVN_3(\mu, \Sigma)$ where $\mu = (5,6,7)$ and
$\Sigma = \begin{bmatrix}2 & 0 & 1\\0 & 3 & 2\\1&2&4\end{bmatrix}$
(a) The marginal distribution of $Y_1$
(b) The joint distribution of $Y_3$ and $Y_1$
(c) The conditional distribution of $Y_3$ given $Y_1$.
Now, my experience with probability prior to this class is limited to Sheldon Ross's intro book and Wackerley, Mendenhall, and Schaeffer, which I'm currently using for my Into Math Stat class. I've never before done conditional, joint, or marginal densities with matrices, nor have I covered them for variables that are dependent.... So please forgive the embarassing attempts I have made to solve this. It also does not help that the class this problem is for does not have a textbook, just some notes frantically scrambled onto a whiteboard twice a week while our prof reminds us that he hates teaching and would rather be researching...
My attempts
(a) I can write
$Y_1 = AY + b$, where $b=(0,0,0)^T$ and $A = (1,0,0)^T$.
Thus: $Y_1$~$MVN(\cdot,\cdot)$ because it is a linear transformation of a multivariate normal.
Thus: $E[Y_1]=E[AY+b]=AE[Y]+b=(1,0,0)^T(5,6,7)^T=5$
$Cov(AY) = ACov(Y)A^T = A\begin{bmatrix}2 & 0 & 1\\0 & 3 & 2\\1&2&4\end{bmatrix}A^T$
This becomes: $(2,0,1)\begin{bmatrix}1\\0\\0\end{bmatrix}=2$
However, knowing that $Y_1$ and $Y_3$ are not independent makes me question this result...

(b) The joint distribution of $Y_1$ and $Y_3$
I basically did the same thing as above but with $A = \begin{bmatrix}1 & 0 & 0\\0&0&1\end{bmatrix}$.
This yielded $W = (Y_1, Y_3)^T$ ~ $MVN(\mu=(5,7)^T, \Sigma = \begin{bmatrix}2 & 1\\1&4\end{bmatrix})$
(c) The conditional distribution of $Y_3$ given $Y_1$
This part I'm honestly at a loss for...unless I'm just supposed to take the integral of the density function for $(Y_1, Y_3)$ divided by the density function for $(Y_1)$, i.e., conditional = $\int f_{Y_1,Y_3}/f_{Y_1} dy_1$.
If I am supposed to do this, how do I get the actual joint density function and perform the integration and division?
The only formulas I've been provided for multivariate normal distributions involve matrices, and I have never seen matrix integration. I apologize if my comprehension is lacking, my classmates and I have been thrown into this material with no assistance from our professor.

  • 2
    $\begingroup$ Forget about integration for a moment. You know all these distributions are normal or bivariate normal, so it remains only to find their parameters. For (a), what are the mean and variance of $Y_1$? For (b), what are the means, variances, and covariances of $(Y_1, Y_3)$? (c) is a little harder but can be addressed the same way, understanding that its parameters will be functions of $Y_1$--and they can be figured out from your answer to (b). Or, if you know about ordinary least squares regression, you already know the answer to (c) (perhaps in disguise). $\endgroup$
    – whuber
    Commented Feb 28, 2015 at 0:31
  • $\begingroup$ Hey whuber, thanks for the speedy response! I think I follow you, but I'm not quite sure. $\mu_{Y_1} = 5$, $Var(Y_1) = 2$, $\mu_{Y_1,Y_3} = (5,7)$, $Var(Y_1,Y_3) = \begin{bmatrix}2 & 1\\1&4\end{bmatrix}$, where Cov($Y_1,Y_3$) = 1. $\endgroup$
    – Patrick
    Commented Feb 28, 2015 at 0:33
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    $\begingroup$ That's correct. You've done the easy part :-). Presumably you have already seen how to compute conditional distributions with bivariate Normals, which is what you have now reduced (c) to. $\endgroup$
    – whuber
    Commented Feb 28, 2015 at 0:38
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    $\begingroup$ I have. I believe I would use $E[Y_3|Y_1]=\mu_{Y_3} + Cov(Y_1,Y_3)*(Y_1-\mu_{Y_1}) / Var(Y_1) = \frac{9+Y_1}{2}$ and $V(Y_3 | Y_1) = Var(Y_3) - Cov(Y_1,Y_3)^2 / Var(Y_1) = 7/2$ $\endgroup$
    – Patrick
    Commented Feb 28, 2015 at 0:43
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    $\begingroup$ Now that you have figured out the answer all by yourself (following the hints from @whuber), please write up your work neatly and post it as an answer to your own question. Otherwise, your question will pop up month after month as an unanswered question that needs some attention, and nobody will be inclined to write an answer. $\endgroup$ Commented Feb 28, 2015 at 14:29

1 Answer 1


Alrighty, y'all. I have an answer. Sorry it took me so long to get it posted here. School was absolutely hectic this week. Spring break is here, though, and I can type up my answer.

First we need to find the joint distribution of $(Y_1, Y_3)$. Since $Y\sim MVN( \mu, \Sigma)$ we know that any subset of the components of $Y$ is also $MVN$. Thus we use
$$ A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} $$ And see that
$$ AY = (Y_1, Y_3)^T $$ $$ \Sigma = \begin{pmatrix} 2 & 1 \\ 1 & 4 \\ \end{pmatrix} $$ $$ \mu(Y_1,Y_2) = (5,7)^T $$ Therefore, using the theorem for conditional distributions of a multivariate normal yields:
$$\begin{align}\newcommand{\c}{\text{Cov}} \newcommand{\v}{\text{Var}} E[Y_3|Y_1]&=μ_{Y_3}+\frac{\c(Y_1,Y_3)(Y_1−μ_{Y_1})}{\v(Y_1)}\\ &=\frac{9+Y_1}{2} \end{align}$$ And
$$\begin{align} \v(Y_3|Y_1) &= \v(Y_3) - \frac{\c(Y_1,Y_3)^2}{\v(Y_1)} \\ &= 4 - \frac{1}{2} = \frac{7}{2} \end{align}$$

  • 3
    $\begingroup$ +1 Good job! Thanks for taking the time to write up the answer. $\endgroup$ Commented Mar 7, 2015 at 18:47

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