=\langle \gamma_2, w_\sigma \rangle \\ To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Let $r^{\underline k}$ denote the falling power $r(r-1)(r-2)\dotsm (r-k+1)$. For books, we may refer to these: https://amzn.to/34YNs3W OR https://amzn.to/3x6ufcEThis lecture explains how to find the MGF of Gamma distribution.Gamma D. is then. CRC Standard Mathematical Tables, 28th ed. k is the number of observed successes. M(t)=\frac{_2F_1(-n,-a;b-n+1;e^t)}{_2F_1(-n,-a;b-n+1;1)} Wallenius' noncentral hypergeometric distribution is obtained if n balls are taken one by one. Thread starter Shahab; Start date Mar 24, 2022; S. Shahab Guest. Moment Generating Function (MGF) of Hypergeometric Distribution is No Greater Than MGF of Binomial Distribution with the Same Mean Why are there contradicting price diagrams for the same ETF? f(x) = \frac{\binom{a}{x}\binom{b}{n-x}}{\binom{a+b}{n}} Let $N \in \mathbb{Z}_{>0}.$ Notation: $[N] = \{1,2,,N\}.$ Let $C = \{c_1, , c_N\}$ be a set of variables, and $\mathbb{R}^C$ the vector space of $\mathbb{R}$-linear combinations of $c_j$s. By linearity of expectation, (For $N \in \mathbb{R},$ we use the different definition $N^{\underline{n}} := N (N-1) \cdots (N-n+1)$; take caution not to confuse the two usages. and the multivariate moment generating function can then be written as Intuition Consider a Bernoulli experiment, that is, a random experiment having two possible outcomes: either success or failure. in the covariance summation, Combining equations (), (), (), and () gives the variance, This can also be computed directly from the sum. Now consider $(X_1, X_2, , X_n)$ a random sample from $C$ taken (uniformly and) without replacement. Why should you not leave the inputs of unused gates floating with 74LS series logic? Asking for help, clarification, or responding to other answers. For example, the probability of getting AT MOST 7 black cards in our sample is 0.83808. $Y$ also has mean $\frac{mn}{N}.$. That is, P (X < 7) = 0.83808. $$\mathbb{E}[e^{uX}] = \sum_{k} \frac{{m \choose k} {N-m \choose n-k}}{{N \choose n}} e^{uk}$$ It only takes a minute to sign up. A generic analysis approach referred to as the moment generating function (MGF) method has been introduced for the purpose of simplifying the evaluation of the performance of digital communication over fading channels. Note that the series equation for the confluent hypergeometric function (Kummer's function of the first kind) is. The moment-generating function for Y is m Y(t) := E . When the migration is complete, you will access your Teams at stackoverflowteams.com, and they will no longer appear in the left sidebar on stackoverflow.com. This is because probabilities are never negative. for $k$ running from $0$ to $\min\{n,m\}.$ Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. Thanks for contributing an answer to Mathematics Stack Exchange! Will Nondetection prevent an Alarm spell from triggering? The hypergeometric distribution differs from the binomial distribution in the lack of replacements. The formula follows from the simple fact that E[exp(t(aY +b))] = etbE[e(at)Y]: Proposition 6.1.4. If you continue to use this site we will assume that you are happy with it. By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. With $c(r), s(r),$ and $f^*$ defined as above (see ($\dagger$)), $$\mathbb{E}[f(\mathcal{Y})] = \frac{1}{N^{\underline{n}}} \sum_{x \in C^{\underline{n}}} f^*(x_1,, x_n).$$, Setting $f$ to be a nonzero constant function, we get $$\sum_{(k,r,i)'} \frac{c(r)}{s(r)} = 1$$ Why are standard frequentist hypotheses so uninteresting? Lemma 1 implies $c$ and $s$ are well-defined. The uniqueness property means that, if the mgf exists for a random variable, then there one and only one distribution associated with that mgf. In probability theory and statistics, the hypergeometric distribution is a discrete probability distribution that describes the number of successes in a sequence of n draws from a finite population without replacement, just as the binomial distribution describes the number of successes for draws with replacement. $$ Proof of Theorem 2. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, $$\mathbb{P}[X = k] = \frac{{m \choose k} {N-m \choose n-k}}{{N \choose n}},$$, $$\mathbb{P}[Y = k] = {m \choose k}\left(\frac{n}{N}\right)^k \left(1-\frac{n}{N} \right)^{m-k}.$$, $$(1) \ \ \forall u \in \mathbb{R}, \mathbb{E}[e^{uX}] \leq \mathbb{E}[e^{uY}].$$, $$\forall u \in \mathbb{R}, \mathbb{E}[e^{uX}] \leq \mathbb{E}[e^{uY}].$$, $$\mathbb{E}[e^{uX}] = \sum_{k} \frac{{m \choose k} {N-m \choose n-k}}{{N \choose n}} e^{uk}$$, $$\mathbb{E}[e^{uY}] = \sum_{k} {m \choose k} \left(\frac{n}{N} \right)^k \left(1-\frac{n}{N}\right)^{m-k} e^{uk}.$$, $$(2) \ \ \forall k \leq m,n \leq N, \ \ \frac{{N - m \choose n-k}}{{N \choose n}} \leq \left(\frac{n}{N} \right)^k \left(1 - \frac{n}{N} \right)^{m-k}.$$, $(1), \mathbb{E}[e^{uX}] \leq \mathbb{E}[e^{uY}].$, $e^{uX} = 1 + uX + \frac{u^2 X^2}{2} + \cdots,$, $$(3) \ \ \forall K \in \mathbb{N}, \ \ \mathbb{E}[X^K] \leq \mathbb{E}[Y^K].$$. It is . 3.7 The Hypergeometric Probability Distribution The hypergeometric distribution, the probability of y successes when sampling without15 replacement n items from a population with r successes and N r fail-ures, is p(y) = P (Y = y) = r y N r n y N n , 0 y r, 0 n y N r, We omit it for brevity. Proof. If a box contains N balls, a of them are black and N a are white, and n number of balls are drawn at random without replacement , then the probability of getting x black balls (and obviously n x white balls) is given by the following p.m.f. Therefore, applying Jensen's inequality, $$f(x_1 + \cdots + x_n) \leq \sum_{(k,r,i)'} \frac{c(r)}{s(r)} f(r_1 x_{i_1} + \cdots + r_k x_{i_k}) \\ Moment generating functions (mgf) are a very powerful computational tool. A probability distribution is a mathematical description of the probabilities of events, subsets of the sample space.The sample space, often denoted by , is the set of all possible outcomes of a random phenomenon being observed; it may be any set: a set of real numbers, a set of vectors, a set of arbitrary non-numerical values, etc.For example, the sample space of a coin flip would be . $$\mathbb{E}[f(\mathcal{X})] \leq \mathbb{E}[f(\mathcal{Y})].\ \square$$, Remark 4. a pick- lottery from a reservoir of balls (of which Moment Generating Function (MGF) of Hypergeometric Distribution is No Greater Than MGF of Binomial Distribution with the Same Mean probability combinatorics probability-distributions hypergeometric-function probabilistic-method Comparing these: we have $\frac{n-k}{N-k} \le \frac{n-k+1}{N-k+1} \le \dots \le \frac{n-1}{N-1} \le \frac nN$, so in particular $\frac{n^{\underline k}}{N^{\underline k}} \le \frac{n^k}{N^k}$, and $\mathbb E[X^{\underline k}] \le \mathbb E[Y^{\underline k}]$. Expected value can be negative. Will it have a bad influence on getting a student visa? . Why is there a fake knife on the rack at the end of Knives Out (2019)? How do you find the expected value of two variables? MOMENT GENERATING FUNCTION (mgf) Let X be a rv with cdf F X (x). Can the expected value be greater than 1? MOMENT GENERATING FUNCTION AND IT'S APPLICATIONS 3 4.1. of obtaining correct balls are I currently do not believe this inequality is true, based on some numbers I put into Maple (but I may have made an error there). For examples of the negative binomial distribution, we can alter the geometric examples given in Example 3.4.2. $$, $\frac{n-k}{N-k} \le \frac{n-k+1}{N-k+1} \le \dots \le \frac{n-1}{N-1} \le \frac nN$, $\frac{n^{\underline k}}{N^{\underline k}} \le \frac{n^k}{N^k}$, $\mathbb E[X^{\underline k}] \le \mathbb E[Y^{\underline k}]$, $\mathbb E[e^{uX}] \le \mathbb E[e^{uY}]$, $\mathbb E[e^{v(m-X)}] \le \mathbb E[e^{v(m-Y)}]$, $\mathbb E[e^{-vX}] \le \mathbb E[e^{-vY}]$, $N^{\underline{n}} := N (N-1) \cdots (N-n+1)$, $\# [N]^{\underline{n}} = N^{\underline{n}}$, $$\mathbb{R}^C \supset C_{\oplus n} := \{x_1 + \cdots + x_n : \text{ each } x_i \in C\}.$$, $f : C_{\oplus n} \overset{\sim}{\to} \Gamma$, $y_2 = (y_{2,1},,y_{2,n}) \in C^{\underline{n}}.$, $$\gamma_2 := f(r_1 y_{2, i'_1} + \cdots + r_k y_{2, i'_k}).$$, $\langle \gamma_1, w \rangle = \langle \gamma_2, w \rangle > 0$, $\langle \gamma_1, u \rangle = \langle \gamma_2, u \rangle > 0.$, $\gamma_1 = f(r_1 y_{1,i_1} + \cdots + r_k y_{1,i_k})$, $$w_{\sigma} := N^{-n} \sum_{x \in C^n} f(\sigma(x_1) + \cdots + \sigma(x_n))$$, $$u_{\sigma} := \frac{1}{N^{\underline{n}}} \sum_{x \in C^{\underline{n}}} \left( \sum_{k=1}^n \sum_{\substack{r \in [n]^k, \\ r_1 + \cdots + r_k = n}} \sum_{i \in [n]^k} f(r_1 \sigma(x_{i_1}) + \cdots + r_k \sigma(x_{i_k}))\right),$$, $\sum_{k=1}^n \sum_{\substack{r \in [n]^k, \\ r_1 + \cdots + r_k = n}} \sum_{i \in [n]^k},$, $\gamma_1 = f(r_1 y_{i_1} + \cdots + r_k y_{i_k})$, $$f^* : C^{\underline{n}} \to \mathbb{R}^\Gamma \ \ \ \ (\dagger)$$, $$(x_1, , x_n) \overset{f^*}{\mapsto} \sum_{(k,r,i)'} \frac{c(r)}{s(r)} f(r_1 x_{i_1} + \cdots + r_k x_{i_k}).$$, $$w = \frac{1}{N^{\underline{n}}} \sum_{x \in C^{\underline{n}}} f^*(x_1, , x_n).$$, $\forall l \in \{0,1,,n-1\}, X_{l+1} \in C \setminus \{X_1, , X_l \}$, $$\mathbb{E}[f(\mathcal{X})] \leq \mathbb{E}[f(\mathcal{Y})].$$, $$\mathbb{E}[f(\mathcal{X})] = \frac{1}{N^{\underline{n}}} \sum_{x \in C^{\underline{n}}} f(x_1 + \cdots + x_n)$$, $$\mathbb{E}[f(\mathcal{Y})] = N^{-n} \sum_{y \in C^n} f(y_1 + \cdots + y_n).$$, $$\sum_{(k,r,i)'} \frac{c(r)}{s(r)} = 1.$$, $f(x_1 + \cdots + x_n) = x_1 + \cdots + x_n,$, $$f^*(x_1, , x_n) = \sum_{(k,r,i)'} \frac{c(r)}{s(r)}(r_1 x_{i_1} + \cdots + r_k x_{i_k})$$, $f(x_1 + \cdots + x_n) = x_1 + \cdots + x_n$, $$\mathbb{E}[f(\mathcal{Y})] = \mathbb{E}_{z \in C^{\underline{n}}}[f^*(z)],$$, $$\sum_{(k,r,i)'} \frac{c(r)}{s(r)} = 1$$, $$\sum_{(k,r,i)'} \frac{c(r)}{s(r)} (r_1 x_{i_1} + \cdots + r_k x_{i_k}) = x_1 + \cdots + x_n.$$, $$\mathbb{E}_{x \in C^{\underline{n}}}[f(x_1 + \cdots + x_n)] The mgf has no intrinsic meaning. Then \( N_1 + N_2 \) has the power series distribution relative to the function \( g_1 g_2 \), with parameter value \( \theta \). See below for my complete exposition of this proof. \leq \mathbb{E}_{x \in C^{\underline{n}}}[f^*(x_1, , x_n)],$$, $$\mathbb{E}[f(\mathcal{X})] \leq \mathbb{E}[f(\mathcal{Y})].\ \square$$, Moment Generating Function (MGF) of Hypergeometric Distribution is No Greater Than MGF of Binomial Distribution with the Same Mean, Mobile app infrastructure being decommissioned, Finding the moment generating function of the product of two standard normal distributions, Find the mean of the Geometric distribution from the MGF, Moment Generating Function - Negative Binomial - Alternative Formula, Moment generating function of sample mean and limiting distribution. To learn more, see our tips on writing great answers. Given the above lemma, for any composition $r$ of $n$ (that is, $r \in [n]^k$ and $r_1+ \cdots + r_k = n$), we can define: for any $\gamma_1 = f(r_1 y_{i_1} + \cdots + r_k y_{i_k})$ as in the statement of Lemma 1. The expected value of a constant is just the constant, so for example E(1) = 1.
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