Phân phối xác suất cho các xác suất khác nhau

Nếu tôi muốn có xác suất 9 thành công trong 16 thử nghiệm với mỗi thử nghiệm có xác suất 0,6 thì tôi có thể sử dụng phân phối nhị thức. Tôi có thể sử dụng gì nếu mỗi trong số 16 thử nghiệm có xác suất thành công khác nhau?

Đây là tổng của 16 thử nghiệm nhị thức (có lẽ độc lập). Giả định độc lập cho phép chúng tôi nhân lên xác suất. Vì vậy, sau hai thử nghiệm với xác suất $p_1$ và $p_2$ thành công, cơ hội thành công của cả hai thử nghiệm là $p_1 p_2$ , cơ hội không thành công là $(1-p_1)(1-p_2)$ và cơ hội một thành công là $p_1(1-p_2) + (1-p_1)p_2$ . Biểu hiện cuối cùng đó có giá trị của nó với thực tế là hai cách để có được chính xác một thành công là loại trừ lẫn nhau: nhiều nhất là một trong số chúng thực sự có thể xảy ra. Điều đó có nghĩa là xác suất của họthêm.

Bằng hai quy tắc này - xác suất độc lập nhân và thêm hai loại trừ - bạn có thể tìm ra câu trả lời cho 16 thử nghiệm với xác suất $p_1, \ldots, p_{16}$ . Để làm như vậy, bạn cần tính đến tất cả các cách để có được số lượng thành công nhất định (chẳng hạn như 9). Có $\binom{16}{9} = 11440$cách để đạt được 9 thành công. Một trong số đó, ví dụ, xảy ra khi các thử nghiệm 1, 2, 4, 5, 6, 11, 12, 14 và 15 là thành công và những thử nghiệm khác là thất bại. Những thành công có xác suất$p_1, p_2, p_4, p_5, p_6, p_{11}, p_{12}, p_{14},$và$p_{15}$và những thất bại có xác suất$1-p_3, 1-p_7, \ldots, 1-p_{13}, 1-p_{16}$ . Nhân 16 số này cho cơ hộicủa chuỗi kết quả cụ thể này. Tổng số này cùng với 11.439 số còn lại cho câu trả lời.

Tất nhiên bạn sẽ sử dụng máy tính.

Với nhiều hơn 16 thử nghiệm, cần phải xấp xỉ phân phối. Với điều kiện không có xác suất nào và 1 - p i trở nên quá nhỏ, một xấp xỉ bình thường có xu hướng hoạt động tốt. Với phương pháp này, bạn lưu ý rằng sự mong đợi của tổng n thử nghiệm là μ = p 1 + p 2 + ⋯ + p n và (vì các thử nghiệm độc lập) phương sai là σ 2 = p 1 ( 1 - p 1 ) + $p_i$$1-p_i$$n$$\mu = p_1 + p_2 + \cdots + p_n$ . Sau đó bạn giả vờ sự phân bố của các khoản tiền là bình thường với trung bình μ và độ lệch chuẩn σ . Những câu trả lời có xu hướng tốt cho việc tính toán xác suất tương ứng với một tỷ lệ thành công mà khác với μ bởi không có nhiều hơn một vài bội số của σ . Như n phát triển lớn xấp xỉ này được bao giờ chính xác và công trình hơn cho bội thậm chí lớn hơn của σ xa μ .$\sigma^2 = p_1(1-p_1) + p_2(1-p_2) + \cdots + p_n(1-p_n)$$\mu$$\sigma$$\mu$$\sigma$$n$$\sigma$$\mu$

Một cách khác để xấp xỉ bình thường của @ whuber là sử dụng xác suất "trộn" hoặc mô hình phân cấp. Điều này sẽ được áp dụng khi các là tương tự một cách nào đó, và bạn có thể mô hình này bằng một phân bố xác suất p i ~ D i s t ( θ ) với một hàm mật độ của g ( p | θ ) lập chỉ mục của một số tham số θ . bạn có một phương trình tích phân:$p_i$$p_i\sim Dist(\theta)$$g(p|\theta)$$\theta$

$$Pr(s=9|n=16,\theta)={16 \choose 9}\int_{0}^{1} p^{9}(1-p)^{7}g(p|\theta)dp$$

The binomial probability comes from setting $g(p|\theta)=\delta(p-\theta)$, the normal approximation comes from (I think) setting $g(p|\theta)=g(p|\mu,\sigma)=\frac{1}{\sigma}\phi\left(\frac{p-\mu}{\sigma}\right)$ (with $\mu$ and $\sigma$ as defined in @whuber's answer) and then noting the "tails" of this PDF fall off sharply around the peak.

You could also use a beta distribution, which would lead to a simple analytic form, and which need not suffer from the "small p" problem that the normal approximation does - as beta is quite flexible. Using a $beta(\alpha,\beta)$ distribution with $\alpha,\beta$ set by the solutions to the following equations (this is the "mimimum KL divergence" estimates):

$$\psi(\alpha)-\psi(\alpha+\beta)=\frac{1}{n}\sum_{i=1}^{n}log[p_{i}]$$ $$\psi(\beta)-\psi(\alpha+\beta)=\frac{1}{n}\sum_{i=1}^{n}log[1-p_{i}]$$

Where $\psi(.)$ is the digamma function - closely related to harmonic series.

We get the "beta-binomial" compound distribution:

$${16 \choose 9}\frac{1}{B(\alpha,\beta)}\int_{0}^{1} p^{9+\alpha-1}(1-p)^{7+\beta-1}dp ={16 \choose 9}\frac{B(\alpha+9,\beta+7)}{B(\alpha,\beta)}$$

This distribution converges towards a normal distribution in the case that @whuber points out - but should give reasonable answers for small $n$ and skewed $p_i$ - but not for multimodal $p_i$, as beta distribution only has one peak. But you can easily fix this, by simply using $M$ beta distributions for the $M$ modes. You break up the integral from $0<p<1$ into $M$ pieces so that each piece has a unique mode (and enough data to estimate parameters), and fit a beta distribution within each piece. then add up the results, noting that making the change of variables $p=\frac{x-L}{U-L}$ for $L<x<U$ the beta integral transforms to:

$$B(\alpha,\beta)=\int_{L}^{U}\frac{(x-L)^{\alpha-1}(U-x)^{\beta-1}}{(U-L)^{\alpha+\beta-1}}dx$$

Let $X_i$ ~ $Bernoulli(p_i)$ with probability generating function (pgf):

$$\text{pgf} = E[t^{X_i}] = 1 - p_i (1-t)$$

Let $S = \sum_{i=1}^n X_i$ denote the sum of $n$ such independent random variables. Then, the pgf for the sum $S$ of $n=16$ such variables is:

$$\begin{align*}\displaystyle \text{pgfS} &= E[t^S] \\&= E[t^{X_1}] E[t^{X_2}] \dots E[t^{X_{16}}] \text{ (... by independence)} \\ &= \prod _{i=1}^{16} \left(1-p_i(1-t) \right)\end{align*}$$

We seek $P(S=9)$, which is:

$$\frac{1}{9!}\frac{d^9 \text{pgfS}}{dt^9}|_{t=0}$$

ALL DONE. This produces the exact symbolic solution as a function of the $p_i$. The answer is rather long to print on screen, but it is entirely tractable, and takes less than $\frac{1}{100}$th of a second to evaluate using Mathematica on my computer.

Examples

If $p_i = \frac{i}{17}, i= 1 \text{ to } 16$, then: $P(S=9) = \frac{9647941854334808184}{48661191875666868481} = 0.198268 \dots$

If $p_i = \frac{\sqrt{i}}{17}, i= 1 \text{ to } 16$, then: $P(S=9) = 0.000228613 \dots$

More than 16 trials?

With more than 16 trials, there is no need to approximate the distribution. The above exact method works just as easily for examples with say $n = 50$ or $n = 100$. For instance, when $n = 50$, it takes less than $\frac{1}{10}$th of second to evaluate the entire pmf (i.e. at every value $s = 0, 1, \dots, 50$) using the code below.

Mathematica code

Given a vector of $p_i$ values, say:

n = 16;   pvals = Table[Subscript[p, i] -> i/(n+1), {i, n}];

... here is some Mathematica code to do everything required:

pgfS = Expand[ Product[1-(1-t)Subscript[p,i], {i, n}] /. pvals];
D[pgfS, {t, 9}]/9! /. t -> 0  // N

0.198268

To derive the entire pmf:

Table[D[pgfS, {t,s}]/s! /. t -> 0 // N, {s, 0, n}]

... or use the even neater and faster (thanks to a suggestion from Ray Koopman below):

CoefficientList[pgfS, t] // N

For an example with $n = 1000$, it takes just 1 second to calculate pgfS, and then 0.002 seconds to derive the entire pmf using CoefficientList, so it is extremely efficient.

@wolfies comment, and my attempt at a response to it revealed an important problem with my other answer, which I will discuss later.

Specific Case (n=16)

There is a fairly efficient way to code up the full distribution by using the "trick" of using base 2 (binary) numbers in the calculation. It only requires 4 lines of R code to get the full distribution of $Y=\sum_{i=1}^{n} Z_i$ where $Pr(Z_i=1)=p_i$. Basically, there are a total of $2^n$ choices of the vector $z=(z_1,\dots,z_n)$ that the binary variables $Z_i$ could take. Now suppose we number each distinct choice from $1$ up to $2^n$. This on its own is nothing special, but now suppose that we represent the "choice number" using base 2 arithmetic. Now take $n=3$ so I can write down all the choices so there are $2^3=8$ choices. Then $1,2,3,4,5,6,7,8$ in "ordinary numbers" becomes $1,10,11,100,101,110,111,1000$ in "binary numbers". Now suppose we write these as four digit numbers, then we have $0001,0010,0011,0100,0101,0110,0111,1000$. Now look at the last $3$ digits of each number - $001$ can be thought of as $(Z_1=0,Z_2=0,Z_3=1)\implies Y=1$, etc. Counting in binary form provides an efficient way to organise the summation. Fortunately, there is an R function which can do this binary conversion for us, called intToBits(x) and we convert the raw binary form into a numeric via as.numeric(intToBits(x)), then we will get a vector with $32$ elements, each element being the digit of the base 2 version of our number (read from right to left, not left to right). Using this trick combined with some other R vectorisations, we can calculate the probability that $y=9$ in 4 lines of R code:

exact_calc <- function(y,p){
    n       <- length(p)
    z       <- t(matrix(as.numeric(intToBits(1:2^n)),ncol=2^n))[,1:n] #don't need columns n+1,...,32 as these are always 0
    pz      <- z%*%log(p/(1-p))+sum(log(1-p))
    ydist   <- rowsum(exp(pz),rowSums(z))
    return(ydist[y+1])
}

Plugging in the uniform case $p_i^{(1)}=\frac{i}{17}$ and the sqrt root case $p_i^{(2)}=\frac{\sqrt{i}}{17}$ gives a full distribution for y as:

$$\begin{array}{c|c}y & Pr(Y=y|p_i=\frac{i}{17}) & Pr(Y=y|p_i=\frac{\sqrt{i}}{17})\\ \hline 0 & 0.0000 & 0.0558 \\ 1 & 0.0000 & 0.1784 \\ 2 & 0.0003 & 0.2652 \\ 3 & 0.0026 & 0.2430 \\ 4 & 0.0139 & 0.1536 \\ 5 & 0.0491 & 0.0710 \\ 6 & 0.1181 & 0.0248 \\ 7 & 0.1983 & 0.0067 \\ 8 & 0.2353 & 0.0014 \\ 9 & 0.1983 & 0.0002 \\ 10 & 0.1181 & 0.0000 \\ 11 & 0.0491 & 0.0000 \\ 12 & 0.0139 & 0.0000 \\ 13 & 0.0026 & 0.0000 \\ 14 & 0.0003 & 0.0000 \\ 15 & 0.0000 & 0.0000 \\ 16 & 0.0000 & 0.0000 \\ \end{array}$$

So for the specific problem of $y$ successes in $16$ trials, the exact calculations are straight-forward. This also works for a number of probabilities up to about $n=20$ - beyond that you are likely to start to run into memory problems, and different computing tricks are needed.

Note that by applying my suggested "beta distribution" we get parameter estimates of $\alpha=\beta=1.3206$ and this gives a probability estimate that is nearly uniform in $y$, giving an approximate value of $pr(y=9)=0.06799\approx\frac{1}{17}$. This seems strange given that a density of a beta distribution with $\alpha=\beta=1.3206$ closely approximates the histogram of the $p_i$ values. What went wrong?

General Case

I will now discuss the more general case, and why my simple beta approximation failed. Basically, by writing $(y|n,p)\sim Binom(n,p)$ and then mixing over $p$ with another distribution $p\sim f(\theta)$ is actually making an important assumption - that we can approximate the actual probability with a single binomial probability - the only problem that remains is which value of $p$ to use. One way to see this is to use the mixing density which is discrete uniform over the actual $p_i$. So we replace the beta distribution $p\sim Beta(a,b)$ with a discrete density of $p\sim \sum_{i=1}^{16}w_i\delta(p-p_i)$. Then using the mixing approximation can be expressed in words as choose a $p_i$ value with probability $w_i$, and assume all bernoulli trials have this probability. Clearly, for such an approximation to work well, most of the $p_i$ values should be similar to each other. This basically means that for @wolfies uniform distribution of values, $p_i=\frac{i}{17}$ results in a woefully bad approximation when using the beta mixing distribution. This also explains why the approximation is much better for $p_i=\frac{\sqrt{i}}{17}$ - they are less spread out.

The mixing then uses the observed $p_i$ to average over all possible choices of a single $p$. Now because "mixing" is like a weighted average, it cannot possibly do any better than using the single best $p$. So if the $p_i$ are sufficiently spread out, there can be no single $p$ that could provide a good approximation to all $p_i$.

One thing I did say in my other answer was that it may be better to use a mixture of beta distributions over a restricted range - but this still won't help here because this is still mixing over a single $p$. What makes more sense is split the interval $(0,1)$ up into pieces and have a binomial within each piece. For example, we could choose $(0,0.1,0.2,\dots,0.9,1)$ as our splits and fit nine binomials within each $0.1$ range of probability. Basically, within each split, we would fit a simple approximation, such as using a binomial with probability equal to the average of the $p_i$ in that range. If we make the intervals small enough, the approximation becomes arbitrarily good. But note that all this does is leave us with having to deal with a sum of indpendent binomial trials with different probabilities, instead of Bernoulli trials. However, the previous part to this answer showed that we can do the exact calculations provided that the number of binomials is sufficiently small, say 10-15 or so.

To extend the bernoulli-based answer to a binomial-based one, we simply "re-interpret" what the $Z_i$ variables are. We simply state that $Z_i=I(X_i>0)$ - this reduces to the original bernoulli-based $Z_i$ but now says which binomials the successes are coming from. So the case $(Z_1=0,Z_2=0,Z_3=1)$ now means that all the "successes" come from the third binomial, and none from the first two.

Note that this is still "exponential" in that the number of calculations is something like $k^g$ where $g$ is the number of binomials, and $k$ is the group size - so you have $Y\approx\sum_{j=1}^{g}X_j$ where $X_j\sim Bin(k,p_j)$. But this is better than the $2^{gk}$ that you'd be dealing with by using bernoulli random variables. For example, suppose we split the $n=16$ probabilities into $g=4$ groups with $k=4$ probabilities in each group. This gives $4^4=256$ calculations, compared to $2^{16}=65536$

By choosing $g=10$ groups, and noting that the limit was about $n=20$ which is about $10^7$ cells, we can effectively use this method to increase the maximum $n$ to $n=50$.

If we make a cruder approximation, by lowering $g$, we will increase the "feasible" size for $n$. $g=5$ means that you can have an effective $n$ of about $125$. Beyond this the normal approximation should be extremely accurate.

The (in general intractable) pmf is

$$\Pr(S=k) = \sum_{\substack{A\subset\{1,\dots,n\}\\ |A|=k}} \left( \prod_{i\in A} p_i \right)\left(\prod_{j\in \{1,\dots,n\}\setminus A} (1-p_j) \right) \, .$$ R code:

p <- seq(1, 16) / 17
cat(p, "\n")
n <- length(p)
k <- 9
S <- seq(1, n)
A <- combn(S, k)
pr <- 0
for (i in 1:choose(n, k)) {
    pr <- pr + exp(sum(log(p[A[,i]])) + sum(log(1 - p[setdiff(S, A[,i])])))
}
cat("Pr(S = ", k, ") = ", pr, "\n", sep = "")

For the $p_i$'s used in wolfies answer, we have:

Pr(S = 9) = 0.1982677

When $n$ grows, use a convolution.