Hot answers tagged

46

Start with no children repeat step { Every couple who is still having children has a child. Half the couples have males and half the couples have females. Those couples that have females stop having children } At each step you get an even number of males and females and the number of couples having children reduces by half (ie those that had females ...


38

Let $X$ be the number of boys in a family. As soon as they have a girl, they stop, so \begin{array}{| l |l | } \hline X=0 & \text{if the first child was a girl}\\ X=1 & \text{if the first child was a boy and the second was a girl}\\ X=2 & \text{if the first two children were boys and the third was a girl}\\ \text{and ...


25

I would like to offer a very simple, intuitive explanation. It amounts to looking at a picture: the rest of this post explains the picture and draws conclusions from it. Here is what it comes down to: when there is a "probability mass" concentrated near $X=0$, there will be too much probability near $1/X\approx \pm \infty$, causing its expectation to be ...


21

Summary The simple model that all births independently have a 50% chance of being girls is unrealistic and, as it turns out, exceptional. As soon as we consider the consequences of variation in outcomes among the population, the answer is that the girl:boy ratio can be any value not exceeding 1:1. (In reality it likely still would be close to 1:1, but ...


17

I'll address the issues relevant to either discrete or continuous possibility: A problem with the description of the mean You have a bounded response. But the model you're fitting isn't bounded, and so can blast right through the bound; some of your fitted values may be impossible, and predicted values eventually must be. The true relationship must ...


16

Stevens' scale typology isn't necessarily some inherent characteristic of the variables, nor even data itself, but of how we treat the information - of what we're using it to mean. In some circumstances, exactly the same value may be considered ratio, interval, ordinal or nominal, depending on what we're doing with it - it's a matter of what meaning we give ...


15

Ratios and inverses are mostly meaningful with nonnegative random variables, so I will assume $X \ge 0$ almost surely. Then, if $X$ is a discrete variable which take on the value zero with positive probability, we will be dividing with zero with a positive probability, which explains why the expectation of $1/X$ will not exist. Now look at the continuous ...


14

The birth of each child is an independent event with P=0.5 for a boy and P=0.5 for a girl. The other details (such as the family decisions) only distract you from this fact. The answer, then, is that the ratio is 1:1. To expound on this: imagine that instead of having children, you're flipping a fair coin (P(heads)=0.5) until you get a "heads". Let's say ...


13

Let random variable $X \sim \text{Uniform}(a,b)$ with pdf $f(x)$: where I have assumed $0<a<b$ (this nests the standard $\text{Uniform}(0,1)$ case). [ Different results will be obtained if say parameter $a<0$, but the procedure is exactly the same. ] Further, let $Y \sim N(\mu, \sigma^2)$, and let $W=1/Y$ with pdf $g(w)$: Then, we seek the pdf of ...


10

For independent $a, d\sim\operatorname{\Gamma}(M,c)$, a remarkable result is that $U=a+d$ and $V=a/(a+d)$ are also independent. In addition, $U\sim\operatorname{\Gamma}(2M, c)$, $V\sim\mathrm{B}(M,M)$. See for example Ch.25, Sec. 2 of Johnson, Kotz, and Balakrishnan's Continuous Univariate Distributions, Volume 2 for some details. Now this says $$Y=\frac{a}...


10

Set $$Z = \frac{X}{X+Y} \implies ZX + ZY = X \implies (1-Z)X = ΖY$$ $$ \implies \frac{Z}{1-Z} = \frac {X} {Y} \equiv C$$ where $C$ is a standard Cauchy r.v. Applying a change of variables $$\frac {\partial C}{\partial Z} = \frac {1}{(1-Z)^2}$$ So $$f_Z(z) = \left|\frac {1}{(1-z)^2}\right|f_C[(z/(1-z)] = \frac {1}{(1-z)^2}\frac 1 {\pi}\frac {1}{1+[z/(1-...


9

This is a property of mathematics, it is actually rare that the order of operations does not matter, e.g. the log of the square root is not the same as the square root of the log (except for a few special cases). We often focus on some of those special cases where due to operations distributing, associating, and commuting (flashbacks to algebra, oh no!) we ...


9

I would like to know why $\frac{1}{N} \sum_i\frac{y_i}{x_i}$ is a worse estimator than average of y divided by average of x. Estimator of what? If you want to estimate $E\left(\frac{y}{x}\right)$, then what you have proposed, $\overline{\;\frac{y}{x}}=\frac{1}{N} \sum_i\frac{y_i}{x_i}$, the sample mean of $\frac{y_i}{x_i}$ is often a dandy estimator (...


9

Not only do distributions of untransformed ratios have odd shapes not matching the assumptions of traditional statistical analysis, but there is no good interpretation of a difference in two ratios. As an aside if you can find an example where the difference in two ratios is meaningful, when the ratios do not represent proportions of a whole, please ...


8

It can happen. For instance, if $X$, $Y$ and $Z$ are independent Rademacher variables, i.e. they can be 1 or -1 with equal probability. In this case $X/Y$ is also Rademacher, so has the same distribution as $Z$, while $YZ$ is Rademacher so has the same distribution as $X$. But it won't happen in general. So long as the means exist, necessary (but not ...


8

The multivariate delta method has a heuristic justification here: https://en.wikipedia.org/wiki/Delta_method#Multivariate_delta_method. For the multivariate delta method you have a specific function $f$ that takes a vector argument which is $p$ dimensional and maps this to a $k$ dimensional space. In the case of a ratio estimator $p=2$ and $k=1$. The ...


8

The expectation is infinite. One way to see this is to condition on $H$. Preliminary changes of variable (merely involving rescaling $H$ and $W$ and then shifting to a new origin) reduce the conditional expectation to a positive constant times a two-dimensional integral of the form $$\mathcal{I}(\lambda)=\iint_{\mathbb{C}}\ \frac{1}{|z|^2} e^{-\lambda |z-...


7

This post elaborates on the answers in the comments to the question. Let $X = (X_1, X_2, \ldots, X_n)$. Fix any $\mathbf{e}_1\in\mathbb{R}^n$ of unit length. Such a vector may always be completed to an orthonormal basis $(\mathbf{e}_1, \mathbf{e}_2, \ldots, \mathbf{e}_n)$ (by means of the Gram-Schmidt process, for instance). This change of basis (from ...


7

The expectation is undefined. Let the $X_i$ be iid according to any distribution $F$ with the following property: there exists a positive number $h$ and a positive $\epsilon$ such that $$F(x) - F(0) \ge h x\tag{1}$$ for all $0 \lt x \lt \epsilon$. This property is true of any continuous distribution, such a Normal distribution, whose density $f$ is ...


6

Couples with exactly one girl and no boys are the most common The reason this all works out is because the probability of the one scenario in which there are more girls is much larger than the scenarios where there are more boys. And the scenarios where there are lots more boys have very low probabilities. The specific way it works itself out is illustrated ...


6

Imagine tossing a fair coin until you observe a head. How many tails do you toss? $P(0 \text{ tails}) = \frac{1}{2}, P(1 \text{ tail}) = (\frac{1}{2})^2, P(2 \text{ tails}) = (\frac{1}{2})^3, ...$ The expected number of tails is easily calculated* to be 1. The number of heads is always 1. * if this is not clear to you, see 'outline of proof' here


6

A $t$-test assumes the relevant data (differences or ratios) are drawn from a normally distributed population. Thus, at most, one of those possibilities affords a valid $t$-test. In addition, a $t$-test of the mean ratio against $0$ makes no sense whatsoever, so you would need to specify some value that makes intrinsic sense given your data and the ...


6

The answer from @FrankHarrell, and associated comments from him and @NickCox, answer the question admirably. I would add that the implicit focus on the shape of raw distributions of predictors and outcome variables is misplaced; in linear modeling, what's important is linearity of relations of predictors to outcome and the distribution of residuals. I also ...


6

Your situation is a case of attempting to meta-analyze dependent effect sizes. I'll outline the various ways one could proceed in answering your first question, and then describe my sense of how one ought to proceed in answering your second question. How You Could Handle Dependent Effect Sizes in Meta-Analysis Cheung (2014) has a nice summary of the ...


6

For two independent "benignly" distributed random variables $X$ and $Y$, the ratio $Z = X/Y$ has the pdf $$p_Z(z) = z^{-2} \int_{-\infty}^\infty |x|\, p_X(x)\, p_Y(x/z)\, dx \sim z^{-2}\, p_Y(0)\, \langle|X|\rangle, \quad |z| \to \infty.$$ This indicates why the tails of the ratio are Cauchy-like given that the denominator $Y$ has a smooth finite pdf at zero....


5

You can also use simulation: p<-0 for (i in 1:10000){ a<-0 while(a != 1){ #Stops when having a girl a<-as.numeric(rbinom(1, 1, 0.5)) #Simulation of a new birth with probability 0.5 p=p+1 #Number of births } } (p-10000)/10000 #Ratio


5

A simple way to do this is to use a test that the proportion is greater than some threshold--for example, that your proportion, $p$, is greater than 0.5. The most basic one sample proportion test is $$z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$$ The $z$ score can then be tested in the usual way to get a $p$-value. For instance, in your example if ...


5

You might find it even more surprising to discover that even with a nice linear operator like expectation, you still have this issue (median is non linear, mean is linear): $$\text{mean}(A)/\text{mean}(B) \neq \text{mean}(A/B)$$ For example: a=1:5 b=6:10 mean(a)/mean(b) [1] 0.375 mean(a/b) [1] 0.3543651 But then you should already know that in ...


5

Mapping this out helped me better see how the ratio of the birth population (assumed to be 1:1) and the ratio of the population of children would both be 1:1. While some families would have multiple boys but only one girl, which initially led me to think there would be more boys than girls, the number of those families would not be greater than 50% and ...


5

If we assume that the underlying normals are jointly normal, then the result is very simple. This answer uses results from A. H. Joarder (2007/2009), Moments of the product and ratio of two correlated chi-square variables (open access). We have $$r\equiv \frac{s_y^2}{s_x^2} = \frac {\sigma^2_y}{\sigma^2_x}\frac{(n-1)s_y^2/\sigma^2_y}{(n-1)s_x^2/\sigma^...


Only top voted, non community-wiki answers of a minimum length are eligible