1
$\begingroup$

I am trying to predict the covariance of two linear combinations of normal random variables: $\newcommand{\N}{\mathcal N}$ \begin{align} X &= w\N(u_1,\sigma^2_1)+(1-w)\N(u_2,\sigma^2_2) \\ Y &= w\N(u_1,\sigma^2_1)+(1-w)\N(u_3,\sigma^2_3) \end{align} where $w$ can range from $0$ to $1$.

I've tried solving for $\text{cov}(X,Y)$ using \begin{align} \text{cov}(X,Y) &= \text{E}(XY) - \text{E}(X)\text{E}(Y) \\ \text{cov}(X,Y) &= \text{corr}(X,Y)\sigma_X\sigma_Y \end{align}

but am not sure how to find $\text{E}(XY)$ in the first case and $\text{corr}(X,Y)$ in the second.

$\endgroup$
7
  • $\begingroup$ Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. $\endgroup$ Commented Jan 8, 2016 at 17:55
  • 1
    $\begingroup$ Thanks for the information. No, this isn't for a course, I'm trying to apply this to a research study. $\endgroup$
    – Matt P
    Commented Jan 8, 2016 at 17:59
  • $\begingroup$ Is your model: $X = w A + (1-w) B$; $Y=w A + (1-w) C$, where A, B, and C are all uncorrelated? If so, $cov(X,Y) = w^2 \sigma^2_1$ . $\endgroup$ Commented Jan 8, 2016 at 18:39
  • $\begingroup$ @MarkL.Stone Could I also ask about the more complicated case, where A, B, and C are not all uncorrelated? E.g., the same overall model for X and Y, but where A = kB + (1-k)C? $\endgroup$
    – Matt P
    Commented Jan 8, 2016 at 21:51
  • $\begingroup$ @Matt P Expand out the covariance into a sum of terms, as per en.wikipedia.org/wiki/Covariance#Properties . Then simplify each term. In your original version, 3 out of 4 additive terms came out to zero. $\endgroup$ Commented Jan 8, 2016 at 21:57

2 Answers 2

1
$\begingroup$

HINT:

\begin{align*} cov(aX+bY, cV + dW) &= E[acXV + adXW + bcYV + bd YW]\\ &-E[aX+bY]E[cV+dW]\\ &= acE[XV]+adE[XW]+ bcE[YV] + bdE[YW]\\ &- acE[X]E[V]-adE[X]E[W]-bcE[Y]E[V]-bdE[Y]E[W]\\ &= ac \times cov(X,V) + ad \times cov(X,W) + bc\times cov(Y,V) + bd \times cov(Y,W) \end{align*}

$\endgroup$
0
$\begingroup$

w is a constant ranging from 0 to 1.

Let P~ N(u1,σ21), Q ~ N(u2, σ22) and R~ N(u3, σ23)]

If variables P, Q, and R are independent, then

Cov (X, Y) = Cov [w*N(u1,σ21)+(1−w)N(u2, σ22), wN(u1,σ21)+(1−w)*N(u3, σ23)]

= Cov [wP + (1-w) Q, w*P + (1-w)*R]

= w2*Cov(P, P) + w*(1-w)Cov(P, R) + + w(1-w)*Cov(Q, P) + (1-w)2*Cov(Q, R)

= w2* σ21 [If variables P, Q, and R are pair-wise independent]

Alternately, by using Cov(X, Y) =E(XY)−E(X)E(Y)

Cov (X, Y) = Cov [wP + (1-w) Q, w*P + (1-w)*R]

= E[w2*P2 + w*(1-w)PR + + w(1-w)*QP + (1-w)2*QR] – E[wP + (1-w) Q]* E[w*P + (1-w)*R]

= w2*E(P2) + w*(1-w)E(PR) + + w(1-w)*E(QP) + (1-w)2*E(QR) -[w*u1 + (1-w)u2] [w*u1 + (1-w)*u3]

= w2*E(P2) - w2*u2

[If variables P, Q, and R are pair-wise independent, then E(XY) = E(X)*E(Y)]

= w2* σ21

Note: if the variables P, Q and R are not independent, then we would require more information regarding the co-variance/correlation between them.

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.