# Tag Info

## New answers tagged binomial-distribution

1

This is along the lines you started to try. $\sum{X_i}$ is binomial with parameters $p$ and $mn$. The likelihood function is $$f(m)$$ $$=\frac{\Gamma(m+n+1)}{\Gamma(m+n-\sum{X_i}+1)\Gamma(\sum{X_i}+1)}p^{\sum{X_i}}(1-p)^{mn-\sum{X_i}}$$ for $\max{X_i}\le m$. Start by treating the likelihood function as if it were a function of a continuous variable $m \in (\... 1 We have the likelihood of$m$by $$f(m)=\frac{m!}{(m-X)!X!}p^X(1-p)^{m-X}$$ Here$m\geq X$. Now, let's consider when$\frac{f(m+1)}{f(m)}$is less than 1. When it starts to be less than 1, we know that function$f$reached a (local) maximum. After calculation we find $$r = \frac{f(m+1)}{f(m)} = \frac{(m+1)(1-p)}{m+1-X},$$ and when$n\geq\left\lfloor\frac{X}{...

1

I don't see how to do this with only one observation, if $p$ is unknown. If $p$ is known, here are some clues. [More generally, this is a much-studied problem; perhaps see this paper and its references.] If $p = 0.3$ and your observation is $X = 12,$ then the method of moments estimator is $X/p = 40.$ Perhaps it is reasonable to guess that the MLE will be ...

3

If you still need a formal reference the following should do: W Feller, An Introduction to Probability Theory and its Applications, volume 1, third edition, chapter 9, section 9: Problems for solution, problem 6. That should also satisfy the reviewer it is not a very difficult problem!

1

You have a binomial distribution $B(n,p)$, where $n$ is the number of trials, and $p$ is the probability of success. For the binomial distribution, the random variable is $K$, the number of successes, and it has expected value $\mu = np$ and variance $\sigma^2 = np(1-p)$. In your example, $n = 50$, and $p$ is unknown. Let $k$ be the number of successes that ...

3

Double-no (it maximises the variance) The answer from whuber is excellent (except that he makes the highly unreaslistic modelling assumption of attributing such obstinant behaviour to the wrong sex!). To supplement that answer, it is also worth examining what happens if you assume that the votes are independent. If we take the votes as mutually independent ...

1

Garrett, can you please clarify how many values of your response variable Pres you have available for each site identified by a SiteID? Do you have (i) multiple values of Pres or (ii) a single value of Pres for that site? The summary for your most complex model includes this line: Number of obs: 115, groups: SiteID:(HUC12_f:HUC4_f), 115; HUC12_f:HUC4_f, 42;...

2

You can use a mixed effects model, a logistic regression with random intercepts for question and for participant. If you have data in long format, something like participant question answer 1 1 A 1 2 B . . . n 1 A n 2 A the logistic regression will have a linear predictor $$\eta_{... 2 Your situation is similar to that of item-response-theory, so look into that. Otherwise your suggestion seems the right one, use a random intercept per subject, and another random intercept per question. 0 I think the coverage is around 93%, slightly under the target 95%. Figure 1, Coverage Probability (CP) against P(10) I assessed it with a simulation. Coverage changes with probability, so it would be good to try it with various values of the probabilities of the four possible outcomes (P11, P10, P01, P00). Unfortunately, running many values of each P would ... 1 The small sample size means that any asymptotic results will not be appropriate. A simple alternative is to do a permutation test which would work as follows: Let (x_i, y_i)=(x_{1,i},\dots,x_{p,i}, y_i) be your dataset, for i=1,2,\dots,n. Let T denote the number of correct predictions of your model. Create a permuted dataset by rearranging the (x_i,... 1 There are better methods for your case: First, there are exact methods for a confidence interval for a binomial p, see the many relevant posts here. Use one such method and get a confidence interval (l, u) for p. Then (l^2, u^2) is a confidence interval for p^2. 0$$P[R=r|R+T=u]=\frac{P[R=r,R+T=u]}{P[R+T=u]}=\frac{P[R=r,T=u-r]}{P[R+T=u]} =\frac{{{N_R}\choose{r}}p^r(1-p)^{N_R-r} {{N_T}\choose{u-r}}p^{u-r}(1-p)^{N_T-u+r}} {\sum_{i=\max(0,u-N_T)}^{\min(u,N_R)} {{N_R}\choose{i}}p^i(1-p)^{N_R-i} {{N_T}\choose{u-i}}p^{u-i}(1-p)^{N_T-u+i}}=\frac{{{N_R}\choose{r}}{{N_T}\choose{u-r}}} {\sum_{i=\max(0,u-N_T)}^{\min(u,...

1

Just try one example and you will find it does not follow a simple form. For example, a X ~ binomail(1, .5) distribution, so the number of successes after one try with 50% probability. That's either one or zero. Now what distribution does aX + b follow? It's either a or a + b, both with 50% probability. There is, as far as I know, no simple formula or name ...

7

The moment generating function is \begin{aligned} \phi(t_1,\ldots, t_k) &= E\left[\exp\left(t_1 X_1 + \cdots + t_k X_k\right)\right]\\ &= \sum_\mathbf{x} \binom{n}{\mathbf x} (p_1 e^{t_1 x_1})\cdots (p_k e^{t_k x_k}) \\ &=\left(p_1 e^{t_1x_1} + \cdots + p_k e^{t_k x_k}\right)^n. \end{aligned} The covariance for $i=1,$ $j=2$ is \begin{...

2

There are several ways to construct confidence intervals for proportion parameters which have better properties than the normal approximation ("Wald-type" confidence interval) which you have mentioned. A relatively simple choice that avoids "overshoot", is to compute a Wald-type confidence interval for the log-transformed proportion and ...

2

You can look at the Wikipedia article on confidence intervals for binomial proportions. About half a dozen methods are in common use. In elementary US statistics books the 95% Wald interval (based on a limiting argument) was used for some years before it was shown to have poor coverage probability for small $n$ (and also to give interval endpoints outside \$(...

1

The problem is that for you are trying to apply the CLT, but you have no control over the rate of convergence. Your assumption that the CLT sufficiently applies so that the confidence interval is MLE +/- 1.96 * SD's is not true if you are getting a CI of (-0.1, 0.3). In fact, it is well known that CLT convergence in the binomial case is quite slow for p ...

0

Definitely, you can and should truncate the CI. It is standard practice. If the un-truncated CI contains the true parameter 95% of the time, then the truncated CI will also contain the true parameter 95% of the time (because you are only removing values that are impossible). Better than truncating the asymptotic CI is to use one of the methods for small ...

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