0
$\begingroup$

We have $10$ cards numbered from $1$ to $10$. We pick two cards among them. What is the expected value of the sum of these two cards ?

I have solved this question the hard way using the law of total expectation (conditioning on the first draw) and I have found that the answer is $11$ (which I checked is right).

But $11=5.5\cdot 2$ so it is equal to the expectation of the sum if we picked the the two cards with replacement (because then the mean of each card would be $5.5$)

So I wonder if there is a hidden argument to get this answer faster ? Thanks !

$\endgroup$
2
  • 1
    $\begingroup$ It might help you to look at my answer here. $\endgroup$
    – Em.
    Jul 13, 2016 at 8:23
  • $\begingroup$ @probablyme thanks this really helped. $\endgroup$
    – Dark
    Jul 13, 2016 at 8:32

2 Answers 2

Reset to default
3
$\begingroup$

The expectation of the sum of two random variables is the sum of their expectations.

$$\mathsf E(X_1+X_2)=\mathsf E(X_1)+\mathsf E(X_2)$$

This is called the Linearity of Expectation.   It works whether the random variables are independent or not.

It is a very useful thing to know.

So whether drawing with or without replacement the expected value is the same.   $\tfrac {11}2+\tfrac {11}2=11$.

From first principles:

$$\begin{align}\mathsf E(X_1+X_2) =&~\begin{cases}\sum_{i=1}^{10}\Big(\tfrac i{10}+\tfrac 1{100}\sum_{j=1}^{10}j\Big) & : \text{with replacement}\\\sum_{i=1}^{10}\Big(\tfrac i{10}+\tfrac 1{90}\mathop{\cdot\sum_{j=1}^{10}}\limits_{j\neq i} j\Big) & : \text{without replacement}\end{cases} \\[1ex]=&~\begin{cases}\tfrac 1{10}\sum_{i=1}^{10}\Big(i+\tfrac 1{10}\frac{110}{2}\Big) & : \text{with replacement}\\\tfrac 1{10}\sum_{i=1}^{10}\Big(i+\tfrac 19(\frac{110}{2}-i)\Big) & : \text{without replacement}\end{cases} \\[1ex]=&~\begin{cases}\tfrac 1{10}\Big(\frac{110}{2}+\frac{110}{2}\Big) & : \text{with replacement}\\\tfrac 1{10}\Big(\tfrac{8}{9}\frac{110}{2}+\tfrac {10}9(\frac{110}{2})\Big) & : \text{without replacement}\end{cases} \\[1ex]=&~\begin{cases}11 & : \text{with replacement}\\11 & : \text{without replacement}\end{cases} \end{align}$$

$\endgroup$
1
  • $\begingroup$ See my comment to @drhab $\endgroup$
    – Dark
    Jul 13, 2016 at 8:24
2
$\begingroup$

If $X_1,X_2$ are the numbers on the two cards drawn then:$$\mathbb E(X_1+X_2)=\mathbb EX_1+\mathbb EX_2=\frac{11}2+\frac{11}2=11$$

Application of linearity of expectation.

$\endgroup$
6
  • $\begingroup$ Thanks. But if you pick $X_1$ first then for $X_2$ you can't pick $X_1$ anymore so I don't see how the expectation of $X_2$ is $\frac{11}2$ $\endgroup$
    – Dark
    Jul 13, 2016 at 8:19
  • $\begingroup$ Do you think that $\mathbb EX_1\neq\mathbb EX_2$ then? $\endgroup$
    – drhab
    Jul 13, 2016 at 8:21
  • $\begingroup$ Because the final result is equal to 11, they must be equal...but I don't see how you can quickly say that $\mathbb EX_2=\frac{11}2$ $\endgroup$
    – Dark
    Jul 13, 2016 at 8:24
  • $\begingroup$ I think of $X_2$ as the number of a card drawn as second from the $10$. But actually the fact that it is drawn as second is not relevant. I could spread out the cards and make up some order in my mind. One of them gets the label "second", but so what? For each card the expectation is the same. $\endgroup$
    – drhab
    Jul 13, 2016 at 8:27
  • $\begingroup$ There is no essential difference between picking out two cards one by one (without replacement and giving them a number) or both at the same time. $\endgroup$
    – drhab
    Jul 13, 2016 at 8:32

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.