Quod Erat Demonstrandum

2009/05/23

點數差的期望值

Filed under: Additional / Applied Mathematics,HKALE,HKCEE,mathematics — johnmayhk @ 3:03 下午

擲兩顆公平骰子一次,求兩個點數差的期望值。

學生甲:

嗯,擲一顆骰子,點數的期望是 \frac{1}{6}(1+2+3+4+5+6) = 3.5。無論擲兩顆,三顆,它們的點數期望值皆是 3.5,故此,兩個點數差的期望值就是 3.5- 3.5= 0

學生乙:

X ,Y 分別是兩顆骰子的點數,則要求的是 E(|X- Y|),即是說當 X \ge Y 時,求的是 E(X - Y) = E(X) - E(Y) = 3.5 - 3.5 = 0,當 X \le Y 時,求的是 E(Y - X) = E(Y) - E(X) = 3.5 - 3.5= 0,即是說,無論哪一種情況,答案也是零。

請問上述學生的答案正確嗎?

讓我返回最初,看看點數差的分佈情況:以縱橫軸表 X,Y 值,坐標表對應的點數差,得下表:

20090523gif01

每坐標出現的概率均等,皆是 \frac{1}{36},經簡單數算,知

P(0) = \frac{6}{36}
P(1) = \frac{10}{36}
P(2) = \frac{8}{36}
P(3) = \frac{6}{36}
P(4) = \frac{4}{36}
P(5) = \frac{2}{36}

其中 P(k) 即點數差是 k 的概率。

故點數差的期望值是 \frac{1}{36}(10\times 1 + 8\times 2 + 6\times 3 + 4\times 4 + 2\times 5) = \frac{35}{18}

看,答案非零!

未運算前,同學會否感到,既然兩骰互相獨立,且點數分佈均勻(各點出現概率皆是 \frac{1}{6}),直觀上,兩點數差的期望值似乎是零,合理吧?學生的算法,問題又出在哪裡?作為授課員,如何透過「非運算」的方式,化解學生的疑問?

5 則迴響 »

  1. E(X-Y) = E(X)-E(Y) = 0;
    E(|X-Y|) = E(X-Y|X>Y)P(X>Y) + E(X-Y|X=Y)P(X=Y) + E(Y-X|X<Y)P(X<Y);
    Since on the RHS, 1st and 3rd term is non-zero trivially and 2nd term is zero, the LHS is non-zero.

    迴響 由 Adrian Tam — 2009/05/23 @ 7:29 下午 | 回覆

  2. Thank you Adrian Tam and thank you for leaving your website:

    http://adrian.idv.hk/doku.php

    the name “Integrable differential" is quite interesting.

    Just another related question to the post:

    Select two values independently and uniformly from [0,1], what is the expected distance between two values? I can use double integral to obtain the value \frac{1}{3}, but can we just consider the symmetry of distribution to obtain the result? Let me explain a bit.

    Suppose two numbers divide [0,1] into three subintervals with lengths l_1, l_2, l_3, is it true to say that l_1, l_2, l_3 have the same distribution? If so, by using the fact that E(l_1) = E(l_2) = E(l_3) and l_1 + l_2 + l_3 = 1, it is easy to obtain E(l_2) = \frac{1}{3}. But how to explain, without advanced mathematics (e.g. double integral something), l_1, l_2 have the same distribution? Or, they actually have different distribution indeed?

    迴響 由 johnmayhk — 2009/05/23 @ 8:14 下午 | 回覆

  3. For the question about [0,1], the answer \frac{1}{3} is correct and the study about similar problem is called “Order Statistics".

    On your specific question, let me try with a simple way:
    Let the first value in [0,1] is X, and the second value is Y. Find Min(X,Y): Min(X,Y)=X iff Y\ge X, which has probability of 1-X. Min(X,Y)=Y is therefore has the probability of X.
    E(Min(X,Y))
    =E[E(Y|Y<X)X+E(X)(1-X)]
    =E[\frac{X}{2}X+X(1-X)]
    =E[X-\frac{1}{2}X^2]
    =\int_0^1 (x-\frac{1}{2}x^2)dx
    =\frac{1}{2}-\frac{1}{6}
    =\frac{1}{3}

    迴響 由 Adrian Tam — 2009/05/24 @ 8:06 上午 | 回覆

  4. Thank you Adrian Tam!

    In your solution, is it true in saying

    Min(X,Y) = E(Y|Y<X)X+E(X)(1-X)?

    Another way to solve the problem is quite traditional, (no double integral of course) let Z = min{X,Y}, it is easy to write down P(Z <= z) and by differentiation, the p.d.f. of Z can be found and hence E(Z) can be calculated. Well, the way is still clumsy. I still wondering that: can the problem be solved 'intuitively' (without calculation)?

    Thank you again!

    迴響 由 johnmayhk — 2009/05/29 @ 5:51 下午 | 回覆

    • I think $min(X,Y)=E(Y|Y<X)X+E(X)(1-X)$ in distribution.

      For "intuitive" method I don't know. But a "heuristic", "qualitative" explanation is the following: Throw $n$ dots into [0,1] randomly, what you expect is that, the $n$ dots placed evenly on the region, so each slice is $\frac{1}{n+1}$. This reasoning is what I learned in the engineering school about noise in wireless communication.

      迴響 由 Adrian Tam — 2009/05/30 @ 4:57 上午 | 回覆


RSS feed for comments on this post. TrackBack URI

發表迴響

在下方填入你的資料或按右方圖示以社群網站登入:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / 變更 )

Twitter picture

You are commenting using your Twitter account. Log Out / 變更 )

Facebook照片

You are commenting using your Facebook account. Log Out / 變更 )

Google+ photo

You are commenting using your Google+ account. Log Out / 變更 )

連結到 %s

在 WordPress.com 建立免費網站或網誌.

%d 位部落客按了讚: