Say $(X_1,X_2,X_3)^T \sim N_3\left(\pmatrix{3\\1\\4}, \pmatrix{6&1&-2\\1&13&4\\-2&4&4} \right)$.

What is the joint pdf of $Y_1$ and $Y_2$ if $Y_1 = 2+X_1+X_2+X_3$, $Y_2 = 5+X_1-X_2+2X_3$?

Let $a=(1,1,1)$, then $Y_1 \sim N(2+a^T \mu,2+a^T\Sigma a) = N(10,31)$, right? Similarly then, we get $Y_2 \sim N(12,11)$.

The joint pdf is then given by

$$ f_{Y_1,Y_2}(y_1,y_2) = \frac{1}{2\pi\times31\times11\times\sqrt{1-p^2}}\times\exp\left(-\frac{1}{2(1-p^2)}\left[\left(\frac{y_1-10}{31}\right)^2+\left(\frac{y_2-12}{11}\right)^2 -2p\left(\frac{y_1-10}{31}\right)\left(\frac{y_2-12}{11}\right)\right]\right)$$

Is this right? What would the correlation $p$ be? 0 because each $X_i$ are independent?

  • $\begingroup$ Those $X_i$ aren’t independent. Your first line gives a non-diagonal covariance matrix. $\endgroup$
    – Dave
    Sep 13, 2020 at 14:37
  • $\begingroup$ @Dave Then i guess i can't calculate $\rho$ and my work is ok then? $\endgroup$
    – CCZ23
    Sep 13, 2020 at 14:40
  • 2
    $\begingroup$ To obtain the joint distribution let $a = \pmatrix{1&1\\1&-1\\1&2}.$ $\endgroup$
    – whuber
    Sep 13, 2020 at 15:00
  • $\begingroup$ @whuber And then $(Y_1,Y_2) \sim N(7 + a^T \mu, 7+ a^T \Sigma a)$ ? The + 2 and +5 confuse me here... $\endgroup$
    – CCZ23
    Sep 13, 2020 at 16:21
  • $\begingroup$ This requires the self-study tag. $\endgroup$ Sep 13, 2020 at 18:29

2 Answers 2


Like @whuber said in the comments, a nice way to proceed is defining the matrix


Such that $(Y_1.Y_2)=(X_1,X_2,X_3)\cdot a+(2,5)$. Now, since this is a constant vector plus a linear transformation of a normally distributed vector, this also has normal distribution. Its mean is found to be $\mu=(2,5)+(3,1,4)\cdot a=(10,15)$. Its variance is given by $a^T\Sigma a=\left[\begin{array}{cc}29&-1\\-1&9\end{array}\right]$.

So, you have $\left[\begin{array}{c}Y_1\\Y_2\end{array}\right]\sim\mathcal N\left(\left[\begin{array}{c}10\\15\end{array}\right], \left[\begin{array}{cc}29&-1\\-1&9\end{array}\right]\right)$.

Hope it was helpful!

  • $\begingroup$ Thanks a lot! Would the correlation coefficient be $\rho_{Y_1,Y_2}=\frac{\text{cov}(Y_1,Y_2)}{\sigma_{Y_1}{\sigma_{Y_2}}} =\pmatrix{-\frac{1}{29\times 9}\\-\frac{1}{29\times 9}}$? $\endgroup$
    – CCZ23
    Sep 14, 2020 at 4:28
  • 1
    $\begingroup$ Yes, the correlation coefficient is -1/(29*9). It's not usually expressed as a vector though: usually you express it as a scalar (just the number -1/(29*9)) or a correlation matrix (which comes in handy when you have more than 2 variables). $\endgroup$
    – PedroSebe
    Sep 14, 2020 at 5:26

One way of finding the joint pdf of $Y = 2+X_1+X_2+X_3$ and $Z = 5+X_1-X_2+2X_3$ without matrix manipulations is to understand that since $X_1, X_2, X_3$ are jointly normal, linear combinations of $X_1, X_2, X_3$ are also jointly normal. Thus, all we need to do is find the means, variances, and covariance of $Y$ and $Z$ and we can write down an explicit formula for their joint pdf that doesn't involve matrices at all.

Do you know about the linearity of expectation and how to use this property to compute \begin{align} \mu_Y &= E[Y] = E[2+X_1+X_2+X_3]\\ \mu_Z &= E[Z] = E[5+X_1-X_2+2X_3]? \end{align}

Do you know how to express the variance of a sum in terms of variances and covariances? That is, do you know how to calculate \begin{align} \sigma_Y^2 &= \operatorname{var}(2+X_1+X_2+X_3)\\ \sigma_Z^2 &= \operatorname{var}(5+X_1-X_2+2X_3) \end{align} in terms of $\sigma_i^2=\operatorname{var}(X_i), i=1,2,3$ and $\operatorname{cov}(X_i, X_j), i, j = 1,2,3, i\neq j$?

Do you know that $\operatorname{cov}(Y,Z)$ can be expressed in terms of $\sigma_i^2=\operatorname{var}(X_i), i=1,2,3$ and $\operatorname{cov}(X_i, X_j), i, j = 1,2,3, i\neq j$ and this allows you to compute the correlation coefficient $\rho_{Y,Z}$?


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