calculating $E[3xy]$ in die throw - mostly about the way you perform the calculation

Question

I don't understand the calculation given in the answers for finding mean of a die throw.

edited because of errata in book: experiment: you throw a fair die. we define X: to be the result of the throw, and also defining Y=0 if the result is even and Y=1 if it's odd

Find: $E[3xy]$.

My attempt:

$E[3xy] = 3E[xy]$. so to find $E[xy]$ I have to calculate $$e[xy]=\sum_x\sum_y xyP(x=i)P(y=j)$$

(is it correct?), or can I calculate it otherwise? so the calculation is, when $(n(S) = 1/6), n(S) \cdot 3 \cdot (1+3+5) \cdots $

and my question is: do we calculate it by the result of a die?

I mean we don't care if its $1$ or $3$, we care that the probability is the same, i.e $1/6$.

what is the difference between the mean of e[x] and e[y]? that we calculate in e[y] only the odd values and e[x] = 3.5?

on a side question, if possible: can we calculate $e[xy]$ using the marginal probability function(using only one variable, $x$ or $y$ for instance)?

would really appreciate an explanation if possible.

thank you

Answer 1

It is not correct to state that $\mathbb E[XY]=\sum_x\sum_yxyP(X=x)P(Y=y)$.

It is correct to state that: $$\mathbb E[XY]=\sum_x\sum_yxyP(X=x\wedge Y=y)$$

Note that $X$ and $Y$ are not independent here. Actually the value taken by $Y$ is completely determined by the value taken by $X$.

The term $xyP(X=x\wedge Y=y)$ takes value $0$ if $x$ is even or if $y=0$

Leaving these terms out then we find:$$\mathbb E[XY]=1\cdot1\cdot P(X=1)+3\cdot1\cdot P(X=3)+5\cdot1\cdot P(X=5)=(1+3+5)\frac16=\frac32$$

So that: $$\mathbb E3XY=3\mathbb E[XY]=\frac92$$

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More directly you could go for:$$\mathbb E[3XY]=3\mathbb E[XY\mid X\text{ is odd})P(X\text{ is odd})+3\mathbb E[XY\mid X\text{ is even})P(X\text{ is even})=$$$$3\mathbb E[X\mid X\text{ is odd})P(X\text{ is odd})+0=3\cdot\frac13(1+3+5)\frac12+0=\frac92$$

Answer 2

$P(X=i) = \frac{1}{6}$ $P(Y=j) = \frac{1}{2}$

$$E[xy]=\sum_x\sum_y xyP(x=i,y=j)$$ $E(3XY) = 3\left(1\times 1\times \frac{1}{6}+3\times 1\times \frac{1}{6}+5\times 1\times \frac{1}{6}\right)$

For X = 2,4,6, Y takes the value of 0 and thus can be omitted in the calcualtion of E(XY)

$$ = 3\left((1+3+5)\times 1 \times\frac{1}{6}\right) = \frac{9}{2}$$

$$E(Y) = \sum_{i=1}^{6} E(y_j=j/x_i=i)P(X=i) = (3\times \frac{1}{6})= \frac{1}{2}$$

by the law of total expectation.

$E(Y_j/X=i) = 1$ for $X_i=1,3,5$ and $E(Y_j/X=i) = 0$ for $X_i = 2,4,6$

$$E(X) = \sum_{i=1}^{6} x_iP(X=i) = \frac{7}{2}$$