Quod Erat Demonstrandum

2010/02/28

最大可能和期望值

Filed under: Additional / Applied Mathematics,HKALE — johnmayhk @ 12:04 上午

在中六的應數堂上隨便說一道習題。

X ~ B(n , p)
P_k = P(X = k) (其中 k = 0, 1, \dots n

求使 P_k 為最大的 k 值。

常法:考慮

P_{k + 1} > P_k
\Leftrightarrow C_{k + 1}^np^{k + 1}(1 - p)^{n - k - 1} > C_{k}^np^{k}(1 - p)^{n - k}
\Leftrightarrow \frac{p}{k + 1} > \frac{1 - p}{n - k}
\Leftrightarrow k < (n + 1)p - 1

即是說

P_0 < P_1 < P_2 < \dots < P_{[(n + 1)p] - 1} < P_{[(n + 1)p]} \ge P_{[(n + 1)p] + 1} \ge \dots \ge P_n

於是,當 k = [(n + 1)p] 時,P_k 最大。

同學提出,可否考慮 E(X) = np

當然,期望值 E(X) = np 與使 P_k 為最大的 k 在基本的意義上不同,但看看

[np][(n + 1)p] 兩值可以是很接近,比如 n 值很大 p 值很小的時候。

還有一些更詳細的考察,但都是停在這兒。

[OT] 之前同學問,如何解以下聯立方程:

2^{x^2 + y} + 2^{x + y^2} = 8 ………… (1)
\sqrt{x} + \sqrt{y} = 2 ……….. (2)

(其中 x, y 為實數)

要多少技巧的,同學不妨試試。

6 則迴響 »

  1. Hi,

    The last question you proposed is a very good one! Thanks for sharing :)

    Cheers,
    Wen Shih

    迴響 由 wenshih — 2010/02/28 @ 8:12 上午 | 回覆

  2. Let a = sqrt{x}, b = sqrt{y}, then we have a, b > 0. (otherwise, it is easy to see that there is no solution)

    Now, we have (8/2)^2 = [(2^{a^4 + b^2} + 2^{a^2 + b^4})/2]^2 >= 2^{a^4 + b^4 + a^2 + b^2} By AM-GM.
    This means, 4 >= a^4 + b^4 + a^2 + b^2.

    Now, by power mean, we have (a^4 + b^4)/2 >= [(a + b)/2]^4 = 1, and likewise, (a^2 + b^2)/2 >= [(a + b)/2]^2 = 1.
    This means, 4 >= a^4 + b^4 + a^2 + b^2 >= 2 + 2 = 4.
    Therefore, by the equality condition of the power means and AM-GM, we have a = b = 1. Done.

    迴響 由 koopakoo — 2010/03/02 @ 2:44 上午 | 回覆

  3. By the way, what does [OT] mean?

    迴響 由 koopakoo — 2010/03/02 @ 2:45 上午 | 回覆

    • Thank you Koopa!

      OT = off topic

      迴響 由 johnmayhk — 2010/03/02 @ 7:48 上午 | 回覆

  4. a symmetric equation, easily we have x=y=1.

    迴響 由 code_vest — 2010/03/18 @ 9:53 下午 | 回覆

    • It seems that the uniqueness of solution cannot be guaranteed by symmetry.

      迴響 由 johnmayhk — 2010/03/18 @ 10:40 下午 | 回覆


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