YTC 發問於 科學及數學數學 · 6 年前

F.5 Maths m1 distribution

It is found that 60% of the cases in a physiotherapy clinic are due to injury. On a

certain day, a physiotherapist treats a number of cases in the clinic.

What is the mean number of cases treated before the first injury case?

The answer in the book is

=E(X) - 1

=(1/0.6) - 1

=0.6667

I want to ask why it is E(X) - 1.

Why the first injury case is equal to 1?

Please help, Thank you!

更新:

RE 知足常樂:

換言之,E(Y) = 1/p - 1只是一個定義,

而當中-1 和 first success 的'first' 相同只是巧合?

另外p=1 時應該不是geometric distribution,

只是舉例吧?

2 個解答

評分
  • 6 年前
    最愛解答

    This is exactly the two definitions of the geometric distribution.

    http://en.wikipedia.org/wiki/Geometric_distributio...

    Let p be the success probability of a (Bernoulli) trial.

    If you define X as the number of trials UNTIL the first success, then

    P(X = x) = (1 - p)^(x - 1) p where x = 1, 2, 3, 4, ... (starts from 1)

    and E(X) = 1/p and Var(X) = (1 - p)/p².

    However, you can also define Y as the number of trials BEFORE the first success, then

    P(Y = y) = (1 - p)^y p where y = 0, 1, 2, 3, ... (starts from 0)

    and E(Y) = 1/p - 1 and Var(Y) = (1 - p)/p².

    Actually, Y = X - 1.

    In the literature, both X and Y are also referred to as being distributed as a geometric random variable.

    Therefore, you need to be careful about the definition.

    2015-05-01 14:31:55 補充:

    Therefore, it is NOT as your claim that "the first injury case is equal to 1", but it is the definition of the geometric random variable.

    I give you an extreme situation then you can understand it easily.

    Use the same question, but change 60% to 100%, that is, p = 1.

    2015-05-01 14:33:13 補充:

    Now, the number of cases treated before the first injury case MUST BE 0, because every case is an injury case.

    Therefore, the mean number is also 0.

    Think about whether it is 1/p or 1/p - 1?

    It should be 1/p - 1 = 1/1 - 1 = 1 - 1 = 0.

    2015-05-01 14:33:46 補充:

    The key issue is therefore, whether it is a number UNTIL the first case, OR it is a number BEFORE the first case.

    2015-05-01 23:10:04 補充:

    回應:

    換言之,E(Y) = 1/p - 1只是一個定義,

    而當中-1 和 first success 的'first' 相同只是巧合?

    > 我嘗試理解你的思維,你以為 -1 是因為有 first 這個字眼。

    > 我不同意。

    > "-1" 是因為 UNTIL 和 BEFORE 剛剛欠了1。

    > 所以,也可以算是你所說的「巧合」,但我寧願你明白以上的 until vs before。

    > 如果是問 second, third, 那些叫 negative binomial distribution

    2015-05-01 23:10:15 補充:

    另外 p = 1 時應該不是geometric distribution,只是舉例吧?

    > 你可以當作「是」,也沒有矛盾。

    > 剛剛是一個極端,而另一個極端是 p = 0,但當然這個 Expectation 是無限。

    2015-05-06 22:22:16 補充:

    你這個 before 其實是 until 來的,你看看題目的第三行:

    ... playing "until" he gets the gift.

    雖然 (b) 寫 before he gets the gift,但你想清楚它是 until (and include) the game which gives him the gift,因此是傳統由 1 開始的 geometric r.v.。

    2015-05-06 22:22:47 補充:

    總之個重點是:

    如果你也計算「成功」那一次,那就是傳統由1開始的 geo. r.v. 即 mean = 1/p。

    如果你只計「成功」之前那些「失敗」次數,那就是由0開始的 geo r.v. 即 mean = 1/p - 1。

    2015-05-07 18:02:47 補充:

    對, 這的確只是剛巧本題的兩件接續事件。

    只要想清楚情況就方便分析了~

  • 6 年前

    TO 知足常樂:

    http://postimg.org/image/wvtp9q9t1/

    我才知道原來 before 不一定是 until 減一,

    不過有種被題目騙了的感覺orz

    2015-05-07 17:54:38 補充:

    那其實(b)是問 the expected amount of money pays until he wins the game,

    只是轉了問法,

    而要 gets the gift 就必須 wins the game,

    所以 before gets the gift 是指 until wins the game?

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