0
$\begingroup$

We throw the symmetric dice $n$ times. Let $X$ - the number of received ones, $Y$ - the number of received sixes. Find $\text{Cov}(X,Y)$.

My solution:

We know that $\text{Cov}(X,Y)= \mathbb{E}(XY)-\mathbb{E}(X)\mathbb{E}(Y)$ and $\mathbb{E}(X)=\mathbb{E}(Y)=\frac{n}{6}$.

How can I find $\mathbb{E}(XY)$?

$\endgroup$
1
$\begingroup$

We first define several new random variables:

  • $X_1, \cdots, X_n$, where $X_i = 1$ if $i$th throw is $1$ and $X_i = 0$ otherwise.
  • $Y_1, \cdots, Y_n$, where $Y_i = 1$ if $i$th throw is $6$ and $Y_i = 0$ otherwise.

Now observe that $$ X = \sum_{i=1}^n X_i \\ Y = \sum_{i=1}^n Y_i $$

And we have \begin{align} &\mathbb{E}[XY] \\ =\ &\mathbb{E}[(X_1 + \cdots + X_n)(Y_1 + \cdots + Y_n)] \\ =\ &\mathbb{E}[\sum_{i=1}^n X_iY_i + \sum_{i=1}^n\sum_{j\neq i}X_iY_j] \\ =\ &\mathbb{E}[\sum_{i=1}^n\sum_{j\neq i}X_i Y_j]\\ =\ &\sum_{i=1}^n\sum_{j\neq i}\mathbb{E}[X_iY_j] \\ =\ &\sum_{i=1}^n\sum_{j\neq i} \frac{1}{36} \\ =\ &\frac{n(n-1)}{36} \end{align}


In the third equality, since $i$th throw can not be $1$ and $6$ at the same time, we must have $X_iY_i = 0$, thus we can ignore those terms here.

$\endgroup$
1
$\begingroup$

Let $X_{i}$ take value $1$ if the $i$-th throw produces a one and value $0$ otherwise.

Let $Y_{i}$ take value $1$ if the $i$-th throw produces a six and value $0$ otherwise.

Then $$XY=\sum_{i=1}^{n}X_{i}\times\sum_{j=1}^{n}Y_{j}=\sum_{i=1}^{n}\sum_{j=1}^{n}X_{i}Y_{j}$$ and consequently:

$$\mathbb{E}XY=\sum_{i=1}^{n}\sum_{j=1}^{n}\mathbb{E}X_{i}Y_{j}$$ If $i\neq j$ then $X_i$ and $Y_j$ are independent, and evidently $X_iY_i=0$.

Can you take it from here?


A more direct route to find $\text{Cov}\left(X,Y\right)$: $$\text{Cov}\left(X,Y\right)=\text{Cov}\left(\sum_{i=1}^{n}X_{i},\sum_{j=1}^{n}Y_{j}\right)=\sum_{i=1}^{n}\sum_{j=1}^{n}\text{Cov}\left(X_{i},Y_{j}\right)$$

If $i\neq j$ then $\text{Cov}(X_i,Y_j)=0$ because $X_i$ and $Y_j$ are independent. So what remains is: $$\text{Cov}\left(X,Y\right)=\sum_{i=1}^{n}\text{Cov}\left(X_{i},Y_{i}\right)=n\text{Cov}\left(X_{1},Y_{1}\right)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.