# Correlation and Covariance exercise - What I did wrong?

#### hellostudent

##### New Member
Hello everyone, I did this exercise:

I have proceeded in this way:
The probability space is initially 7+3=10, So I can compute P(white balls)=0.7 and P(black balls)=0.3.
Then there is the drawing of two balls, So I have computed, through the binomial formula, the probability that the number of white drawn balls was 2,1 or 0:

S

So I have obtained P(x=0)=0.09, P(x=1)=0.42, P(x=2)=0.49
Since then, I've calculated the probability to draw a white or a black ball after the drawing of the precedent 2 balls without replacement:
The probability space becomes 10-2=8, so:
If x=0 (two black ball extracted), the probability to draw a white ball is P(white)=7/8=0.875, and P(black)=0.125
If x=1, P(white)=6/8=0.75, P(black)=0.25;
If x=2, P(white)=5/8=0.625, P(black)=0.375
The, in order to write the joint distribution, I compute: The probability that the first two balls are both black and the third is also black->P(x=0 AND y=0), the probability that the first two balls are black and the third is white->P(x=0 AND y=1) ; The probability that the first two balls are one white and one black and the third is black P(x=1 AND y=0), the probability that the first two balls are one white and one black and the third is white P(x=1 AND y=1), the probability that the first two balls are both white and the third is black P(x=2 AND y=0), the probability that the first two balls are both white and the third is white P(x=2 AND y=1)
To do this, I simply compute the products: P(x=0 AND y=0)=P(x=0)*P(black)=0.09*0.125=0.01125
and so on, until I get this table, that represents the JOINT DISTRIBUTION :

So I compute the Covariance with the formula Cov(X,Y)=E(XY)- μxμy

μx= 0.09*0 + 0.42*1 + 0.49*2 = 1.4
μy= 0.3*0 + 0.7*0 = 0.7
E(XY)= (0*0*0.01125) + (0*1*0.105) + (0*2*0.18375) + (1*0*0.07875) + (1*1*0.315) + (2*1*0.30625) = 0.9275
So Covariance -> Cov(X,Y) = 0.9275 - (1.4*0.7) = -0.0525
Variance of X -:

V(X)=0.42

So V(X)=0.42 and, with the same formula, V(Y)=0.37

Correlation coefficient:

With this formula, the correlation coefficient is ρ(X,Y)= -0.1332

But my exercise sheet results are:
5b) covariance ≈−0.0467; correlation ≈−0.1667.

I have moreover calculated E(X+Y) and I have obtained the same results of my exercise sheet (2.1)

Do you think I did something wrong about covariance and correlation calculation? I know that my results are very similar but I'm not finding any kind of approximation that brings me the results identical to those provided by my professor, so I have a sense that I did something wrong.

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#### Dason

So I didn't get too far into the post. But the marginal distributions for X and Y will not be binomial.

#### hellostudent

##### New Member
So I didn't get too far into the post. But the marginal distributions for X and Y will not be binomial.
You are right! I remade the exercise by using the probability tree and not the binomial formula, considering the first two drawings as dependent one from the other and now the result is correct. Thank you!