Using conditional Variance formula to find conditional variance in terms of X and Y.

Question

The joint and marginal probabilities of X and Y are denoted as X and Y.

1. Calculate the conditional variance of $x$ when $y$ is equal to one. The conditional variance $y$ when $x$ is equal to one.

In order to solve this problem one must find the total of $X$ and $Y$.

In order to do this one must add the values in the chart provided.

So

$$\frac{1}{6} +\frac{1}{3} + \frac{1}{12} = \frac{7}{12}$$

$$\frac{2}{9} +\frac{1}{6} = \frac{7}{18}$$

$$\frac{1}{6} +\frac{2}{9} + \frac{1}{12} = \frac{5}{12}$$

$$\frac{1}{3} +\frac{1}{6} = \frac{1}{2}$$

As a result one gets this chart of the total values.

In order to find the joint probabilities we use this formula.

$\mu_{1}^{'} = E(XY)$

$$ 0 \cdot 0\cdot \frac{1}{6} + 0 \cdot 1 \cdot \frac{2}{9} + 0 \cdot 2 \cdot \frac{1}{36} +1\cdot 0 \cdot \frac{1}{3} +1\cdot 1\cdot\frac{1}{6}+2\cdot0 \cdot \frac{1}{12} = \frac{1}{6}$$

Now in order to find the conditional variance of X and Y. One must use this formula.

$\sigma^2_{X|y} = E[(X-\mu_{X|y})^2|y] = E(X^2|y)-\mu^{2}_{X|y}$

Alas the intricacies of this formula baffles me. How does one use the formula above, in order to derive the conditional variance. Any tips would be whole hardheartedly welcomed!

Answer 1

$\underline {\text{Formulas}}$

• Marginal pmf: $P(Y=y) = \sum_{x\in A} P(X=x,Y=y)$
• Conditional Expec: $E(X|Y=y) = \sum_{x \in A} x \cdot \frac{P(X=x,Y=y)}{P(Y=y)}$
• Conditonal Var : $V(X|Y=y) = E(X^2|Y=y) - E(X|Y=y)^2$

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$$P(Y=1) = \sum^{2}_{x=0} P(X=x,Y=1) \\ P(x=0,y=1)+P(x=1,y=1)+P(x=2,y=1) \\ =\frac{2}{9}+\frac{1}{6} +0 \\ P(Y=1)=\frac{7}{18} \\ E(X|Y=1)=\sum^{2}_{x=0} x \cdot \frac{P(x=x,Y=1)}{\frac{7}{18}} \\ =0 \cdot \frac{P(x=0,Y=1)}{\frac{7}{18}} + 1 \cdot \frac{P(x=1,y=1)}{\frac{7}{18}} + 2 \cdot \frac{P(x=2,y=1)}{(\frac{7}{18})} \\ \frac{\frac{1}{6}}{\frac{7}{18}} \\ E(X|Y=1) =\frac{3}{7} \\ E(x^2|Y=1) = \sum^2_{x=0} x^2 \cdot \frac{P(X=x,Y=1)}{(\frac{7}{18})}\\ \Rightarrow =0^2 \cdot \frac{P(x=0,Y=1)}{\frac{7}{18}} + 1^2 \cdot \frac{P(x=1,y=1)}{\frac{7}{18}} + 2^2 \cdot \frac{P(x=2,y=1)}{\frac{7}{18}} \\ E(x^2|Y=1)= \frac{3}{7} \\ \Rightarrow v(x|Y=1) = E(x^2|Y=1)-E(x|Y=1)^2 \\ = \frac{3}{7}-(\frac{3}{7})^2 \\ \bbox[5px,border:2px solid black] {V(x|Y=1) = \frac{12}{49} \qquad } $$

this should explain/answer your question