Joint Distributions, Covariance and Independence · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Two variables moving together

Mathematics · Probability · ages 19-20
Name ______________________   Date ____________

  1. The joint density is f(x,y)=x+y on 0≤x,y≤1 and zero elsewhere. Find the marginal density value f_X(0.2).

    Answer: ______________

  2. The only joint masses are p(0,0)=0.15, p(0,2)=0.25, p(1,0)=0.35 and p(1,2)=0.25. Here p(x,y)=P(X=x,Y=y). Find P(X=1).

    Answer: ______________

  3. X and Y have finite second moments and positive covariance. What does this sign establish?

    • Increasing X by intervention must increase Y, regardless of any other influences.
    • The probability-weighted mean product of the centered deviations is positive.
    • Every increase in X must be accompanied by an increase in Y, whatever values occur.

  4. Let p(x,y)=P(X=x,Y=y). The only masses are p(0,-1)=0.1, p(1,-1)=0.2, p(0,0)=0.15, p(1,0)=0.25, p(0,2)=0.2 and p(1,2)=0.1. Find P(Y>0).

    Answer: ______________

  5. X and Y each take 0, 1 and 2, with each marginal probability equal to 1/3. P(X=0,Y=0)=1/9 matches its marginal product. This one match proves independence.

    Circle one:   True   False

  6. Cov(X,Y)=-0.4 and X,Y have finite second moments. Let U=3X while Y is unchanged. What is Cov(U,Y)?

    • -0.4, because multiplying one variable never changes covariance.
    • 0.4, because multiplying by a positive constant reverses the covariance sign.
    • -1.2, because the covariance is multiplied by 3.
    • -3.6, because covariance is multiplied by the square of the scale factor.

  7. Lee says independent variables with finite second moments can have nonzero covariance. Is this claim true?

    Circle one:   True   False

  8. Let p(x,y)=P(X=x,Y=y). The only joint masses are p(0,0)=0.3, p(0,3)=0.2, p(2,0)=0.1 and p(2,3)=0.4. Find Cov(X,Y).

    Answer: ______________

  9. The joint density is f(x,y)=x+y on 0≤x,y≤1 and zero elsewhere. Find 144×Cov(X,Y).

    Answer: ______________

  10. X takes -2, 0 and 2 with probabilities 1/4, 1/2 and 1/4, and Y=X². E[X]=E[XY]=0, so Cov(X,Y)=0. Which argument correctly classifies the pair?

    • They are independent because cancellation in E[XY] makes all joint probabilities factor.
    • They are dependent because P(X=0,Y=0)=1/2 differs from the marginal product 1/4.
    • They are dependent because every function of X depends on X, even when that function is constant.
LightMySky · lightmysky.comW1-mt_YyTFTAqhSF-s1

Answer key

For grown-ups. Fold this page away before handing over the rest.

Two variables moving together W1-mt_YyTFTAqhSF-s1

  1. 0.7 · Integrating over y gives f_X(x)=x+1/2 on [0,1]. Therefore f_X(0.2)=0.7.
  2. 0.6 · Add both masses with X=1: 0.35+0.25=0.6.
  3. The probability-weighted mean product of the centered deviations is positive. · Covariance averages products of centered deviations. A positive average does not establish a causal effect or a rule for every possible pair.
  4. 0.3 · Only Y=2 satisfies Y>0. Its masses sum to 0.2+0.1=0.3.
  5. False · Independence requires equality at every pair. One matching cell in this three-by-three table leaves other joint probabilities unchecked.
  6. -1.2, because the covariance is multiplied by 3. · Cov(3X,Y)=3 Cov(X,Y)=3×(-0.4)=-1.2. Only one variable is rescaled.
  7. False · Independence gives E[XY]=E[X]E[Y]. Subtracting these equal quantities gives covariance zero.
  8. 0.6 · E[X]=1, E[Y]=1.8 and E[XY]=2.4. Thus Cov(X,Y)=2.4-1×1.8=0.6.
  9. -1 · E[X]=E[Y]=7/12 and E[XY]=1/3. Covariance is 1/3-49/144=-1/144, so the requested value is -1.
  10. They are dependent because P(X=0,Y=0)=1/2 differs from the marginal product 1/4. · Y=0 occurs exactly when X=0. Both marginal probabilities are 1/2, but their joint probability is 1/2 rather than 1/4.
Worksheet · LightMySky