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It is a pretty simple question, but I wanted to know if I got it right (and I had a weird result in the end):

There are 3 White balls and 1 Green ball in a urn. We take one ball from it, write down the colour then put it back with +c balls of the same colour. Now we take a second ball. $X_{i} = 1$ if the ith ball is green and $X_{i} = 0$ if it is white, for $i = 1,2$.

a) What is the joint probability distribution of $X_{1}$ and $X_{2}$

b) What is the correlation between $X_{1}$ and $X_{2}$. What whappens when c goes to inf?

Here's what I did:

a)

$P(0,0) = \frac{3}{4} \cdot \frac{3+c}{4+c} = \frac{9+3c}{16+4c}$

$P(0,1) = \frac{3}{4} \cdot \frac{1}{4+c} = \frac{3}{16+4c}$

$P(1,0) = \frac{1}{4} \cdot \frac{3}{4+c} = \frac{3}{16+4c}$

$P(1,1) = \frac{1}{4} \cdot \frac{1+c}{4+c} = \frac{1+c}{16+4c}$

b)

$E[X_{1}] = E[(X_{1})^2] = 1 \cdot \frac{1}{4} = \frac{1}{4}$

$E[X_{2}] = E[(X_{1})^2] = 1 \cdot \frac{1+c}{4+c} + 1 \cdot \frac{1}{4+c} = \frac{2+c}{4+c}$ (all other sum terms are 0)

$E[X_{1}X_{2}] = 1 \cdot 1 \cdot \frac{1+c}{4c+16}$ (all other sum terms are 0)

$Var(X_{1}) = \frac{1}{4} - \frac{1}{16} = \frac{3}{16}$

$Var(X_{2}) = \frac{2+c}{4+c} - (\frac{2+c}{4+c})^2 = \frac{4+2c}{(4+c)^2}$

$Cov(X_{1},X_{2}) = \frac{1+c}{4c+16} - \frac{1}{4} \cdot \frac{2+c}{4+c} = -1$

$Cor(X_{1},X_{2}) = \frac{-1}{\sqrt{\frac{3}{16} \cdot \frac{4+2c}{(4+c)^2}}} = \frac{-c-16}{\sqrt{12+6c}}$

And then when c goes to inf, the correlation goes to -inf, which doesn't seem to make sense...

Did I go wrong somewhere?

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1 Answer 1

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I'm having trouble following your logic, but yes, you've made some mistakes (a correlation cannot exceed one in absolute value, for example). $\text{E}(X_1)$ is easy enough to find so let's start by calculating $\text{E}(X_2)$. The key is to condition on $X_1$ and then calculate the expectation in pieces.

\begin{align} \text{E}(X_2) &= \text{E} [ \text{E} (X_2 \mid X_1) ] \\ &= P(X_1 = 1) \text{E} (X_2 \mid X_1 = 1) + P(X_1 = 0) \text{E}(X_2 \mid X_1 = 0) \\ &= \frac{1}{4} \cdot \frac{1 + c}{4 + c} + \frac{3}{4} \cdot \frac{1}{4 + c} \\ &= \frac{4 + c}{4 (4 + c)} \\ &= \frac{1}{4} . \end{align}

This is interesting as it says that on average $X_2$ behaves just like $X_1$. Now since these are Bernoulli random variables with the same expectation the variances are easy:

\begin{align} \text{Var}(X_i) &= \text{E}(X_i^2) - \text{E}(X_i)^2 \\ &= \frac{1}{4} - \frac{1}{16} \\ &= \frac{3}{16} . \end{align}

The only thing left to calculate is the covariance and we can use the identity $\text{Cov}(X_1, X_2) = \text{E}(X_1 X_2) - \text{E}(X_1) \text{E}(X_2)$. We already know the rightmost term so for the other we have

\begin{align} \text{E}(X_2 X_2) &= P(X_1 = 1 \cap X_2 = 1) \\ &= P(X_1 = 1) P(X_2 = 1 \mid X_1 = 1) \\ &= \frac{1 + c}{4 (4 + c)} \end{align}

yielding

\begin{align} \text{Cov}(X_1, X_2) &= \frac{1 + c}{4 (4 + c)} - \frac{1}{16} \\ &= \frac{3c}{16 (4 + c)} . \end{align}

If we now divide by this $\sqrt{\text{Var}(X_1) \text{Var}(X_2)}$ we get

\begin{align} \text{Corr}(X_1, X_2) &= \frac{c}{4 + c} . \end{align}

This makes sense since as $c \to \infty$ we have $X_1 = X_2$ with high probability so the correlation should approach one.

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  • $\begingroup$ Thanks! It seems I forgot that I needed to use the conditional probability when calculating $E[X_{2}]$. Also, is just finding all possible $P(i,j)$ considered the joint probability distribution? $\endgroup$ Sep 28, 2016 at 12:28
  • $\begingroup$ Yes, just enumerating all possible combinations and finding their probabilities is the way to get the joint distribution here. $\endgroup$
    – dsaxton
    Sep 28, 2016 at 13:43

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