57
There is a difference between
independent events: $\mathbb P(A \cap B) =\mathbb P(A)\,\mathbb P(B)$, i.e. $\mathbb P(A \mid B)= \mathbb P(A)$ so knowing one happened gives no information about whether the other happened
mutually disjoint events: $\mathbb P(A \cap B) = 0$, i.e. $\mathbb P(A \mid B)= 0$ so knowing one happened means the other did not happen
...
13
As stated the problem does not make sense, because a joint distribution cannot be found from the marginal distributions! The only meaningful case (as an homework) is to assume independence. In which case the density of the joint distribution is obviously the product of both densities...
13
Not quite, if we use the law of total expectation we would have that
$$
E(X) = E(X| Y \le a)P(Y \le a) + E(X|Y > a) P(Y>a)
$$
13
By definition, the random variables $X_1,\dots,X_n$ are independent iff
$$
\Pr(X_1\in B_1,\dots,X_n\in B_n) = \Pr(X_1\in B_1)\dots\Pr(X_n\in B_n)
$$
for every choice of Borel sets $B_1,\dots,B_n$. Hence, picking $B_i=(-\infty,t_i]$, we have
$$
\Pr(X_1\leq t_1,\dots,X_n\leq t_n) = \Pr(X_1\leq t_1)\dots\Pr(X_n\leq t_n). \qquad (*)
$$
If each $X_i$ has a ...
13
What I understood from your question, is that you might have confused independent events with disjoint events.
disjoint events: Two events are called disjoint or mutually exclusive if they cannot both happen. For instance, if we roll a die, the outcomes 1 and 2 are disjoint since they cannot both occur. On the other hand, the outcomes 1 and “rolling an odd ...
12
A stationary Gaussian process is completely characterized by the combination of its mean, variance and autocorrelation function. The statement as you read it is not true. You need the following additional conditions:
The process is stationary
the process is Gaussian
the mean $μ$ is specified
Then the entire stochastic process is completely characterized ...
answered Jun 19 '12 at 18:00
Michael R. Chernick
37.2k9 gold badges64 silver badges136 bronze badges
12
You actually had your answer right there.
$P(H=hit)$ is the marginal probability. It reads "The probability of getting hit.". It is the proportion of people that got hit crossing the street, irrespective of traffic light.
$P(H=hit|L=red)$ is the conditional probability. It reads "The probability that you get hit, given that the light is red". It is the ...
11
There are many papers addressing such questions.
A good starting place is probably:
Pyke R. (1965), Spacings
Journal of the Royal Statistical Society. Series B (Methodological)
Vol. 27, No. 3 (1965), pp. 395-449
(It has a lot on the continuous case. Many papers refer to this paper, including some that do more with the discrete case.)
You should be ...
answered Jan 18 '13 at 5:05
Glen_b -Reinstate Monica
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11
Here's a counter-example:
Let $X$, $Y$, $Z$ be independent standard normal, and let $W = |Z|\cdot \text{sign}(XY)$.
Then $(W,X)$, $(W,Y)$ and $(X,Y)$ are bivariate normal, but $(W,X,Y)$ is not trivariate normal, since $WXY$ is never negative.
What's happening is that the trivariate distribution has been constructed so that probability is only in four of ...
answered Jan 7 '14 at 9:26
Glen_b -Reinstate Monica
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11
As a concrete example, suppose I toss a coin and roll a die one after the other. As you know, there is a probability distribution associated with the outcomes of both (discrete uniform distributions, i.e. every possible outcome is equally likely). I can use these to find e.g. P(coin lands heads) or P(I roll a six).
The joint probability distribution is just ...
10
Here is a guide to solving this problem (and others like it). I use simulated values to illustrate, so let's begin by simulating a large number of independent realizations from the distribution with density $f$. (All the code in this answer is written in R.)
n <- 4e4 # Number of trials in the simulation
x <- matrix(pmax(runif(n*3), runif(n*3)), nrow=...
9
Yes, it's well-defined. For convenience and ease of exposition I'm changing your notation to use two standard normal distributed random variables $X$ and $Y$ in place of your $X1$ and $X2$. I.e., $X = (X1 - \mu_1)/\sigma_1$ and $Y = (X2 - \mu_2)/\sigma_2$. To standardize you subtract the mean and then divide by the standard deviation (in your post you only ...
9
If I am understanding the passage and table correctly, there are essentially two answers: mathematical and qualitative.
Qualitatively, the "different aspects of forecast quality" is essentially the idea of false positives vs. false negatives, or precision vs. recall. (I will come back to this.)
Mathematically, the table is comparing two binary variables
$$...
9
Presumably $w_1+w_2=1$ and $w_1,w_2 \geq 0$, so $f$ is a convex combination of $f_1,f_2$ and therefore a valid distribution (a mixture of $f_1,f_2$).
Generate a Bernoulli($w_1$) random variable (i.e. flip a coin, heads = 1 with probability $w_1$). If it's $1$, draw a sample from $f_1$. If its $0$, draw a sample from $f_2$. You can use the law of total ...
9
It takes $3\times 2 \times 2 \times 3 = 36$ numbers to write down a probability distribution on all possible values of these variables. They are redundant, because they must sum to $1$. Therefore the number of (functionally independent) parameters is $35$.
If you need more convincing (that was a rather hand-waving argument), read on.
By definition, a ...
8
The joint density uniquely determines the marginal densities:
$$p(x) = \sum_y p(x,y) ~~\text{or}~~
p(x) = \int_{-\infty}^{\infty}p(x,y)\,\mathrm dy$$
(similarly for $p(y)$) and so the conditional densities are also
determined uniquely by the joint density.
So the answer is
No, you cannot construct another group of samples as you desire.
If the ...
8
As you correctly pointed out in your question $f_{Y}(y)$ is calculated by integrating the joint density, $f_{X,Y}(x,y)$ with respect to X. The critical part here is identifying the area on which you integrate. You have already clearly showed graphically the support of the joint distribution function $f_{X,Y}(x,y)$. So, now, you can note that the range of $X$ ...
8
Yes, that's the necessary and sufficient condition for independence not only for two random variables but also for a (finite) sequence of random variables. Check out for example P.2 on page 242 of Probability with Statistical Applications, By Rinaldo B. Schinazi. Or page 259 of
Econometric Analysis of Count Data which is based on probability generating ...
8
The question asks for the expected time to complete both of two independent tasks. Call these times $X_1$ and $X_2$: they are random variables supported on $[0,\infty)$.
Let $F_i$ be the cumulative distribution functions (CDF) of the $X_i$:
$$F_i(x) = \Pr(X_i\le x).$$
The time to complete both tasks is $Y =\max(X_1,X_2)$. Its CDF is given by
$$\eqalign{...
8
Those distributions you call "marginal" are not marginal. They are conditional distributions because you wrote $x \mid y$. The marginal distribution of $X$, for example, is necessarily independent of the value of $Y$.
To see how the conditional distribution is gamma, all you have to do is write $$f_{X \mid Y}(x) = \frac{f_{X,Y}(x,y)}{f_Y(y)} \propto f_{X,...
8
The probability that $X_1=X_2$ is the probability that both are zero, plus the probability that both are one, plus the probability that both are two, and so on.
8
There isn't a "the" with respect to moments, since there are many of them, but moments of bivariate variables are indexed by two indices, not one.
So rather than $k$-th moment, $\mu_k$ you have $(j,k)$-th moments, $\mu_{j,k}$ (sometimes written $\mu_{jk}$ when that's not ambiguous). We might speak of $\mu_{1,1}$, the $(1,1)$ moment or $\mu_{1,2}$, the $(1,2)...
answered Dec 28 '15 at 2:55
Glen_b -Reinstate Monica
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7
$(X,Y)$ take on value $(0,2)$ and $(2,0)$ with equal probability $\frac{1}{2}$. Then,
$$E[X] = 0\times\frac{1}{2} + 2\times\frac{1}{2} = 1,$$
and similarly $E[Y]=1$, while
$$E[XY] = 2\times 0\times\frac{1}{2} + 0\times 2\times\frac{1}{2} = 0.$$
Added note: The $X$ and $Y$ in this example are not uncorrelated random variables
(cf. the title of the question) ...
7
$X$ is not jointly continuous with itself in the sense that there is no joint density
function (pdf) $f_{X,X}(s,t)$ that has positive value over a region of positive area
in the plane
with coordinate axes $s$ and $t$. All the probability mass lies on the straight line of slope $1$ through the origin (a region of zero area) and the joint cumulative
...
7
Yes, it's true that these assumptions imply $X$ and $Y$ are independent.
Simplify the notation by writing $F = F_{X,Y}$. By definition,
$$F(x,y) = \Pr(X \le x, Y \le y).$$
Therefore the limit of $F(x,y)$ as $y$ increases without bound exists and is the chance that $X$ does not exceed $x$:
$$F_X(x) = \Pr(X \le x) = \lim_{y\to\infty} F(x,y) = G_1(x) \lim_{...
6
There are two arguments you need to make. First, if $A$ and $B$ are independent
random variables, then so are $g(A)$ and $h(B)$ independent random variables for
all (measurable) functions $g(\cdot)$ and $h(\cdot)$. So, the question
can be transformed into
For what values of $d$ are $X+dY$ and $X-dY$ independent random variables?
Second, since $X$ and $...
6
The short answer as I understand your q is "yes, but ..." the rates of convergence on S, T, and any other moments are not necessarily the same -- check out determining bounds with the Berry-Esseen Theorem.
In case I misunderstand your q, Sn and Tn even hold to the CLT under conditions of weak dependence (mixing): check out Wikipedia's CLT for dependent ...
6
For McKay's distribution $X$ is a Gamma variate that is the sum of a subset of squares taken from the other, $Y$, which is the sum of a larger set of squares. Implying that $Y>X$ with probability $1$. See McKay's original paper:
McKay, A. T. (1934) Sampling from batches. Journal of the Royal Statistical Society—Supplement 1: 207–216.
6
You might want to check out Karlis and Meligkotsidou, "Multivariate poisson regression with covariance structure". 2005. This paper is about the authors' attempts to model multivariate Poisson variables, which they acknowledge to be a difficult task.
Use of the Mahalanobis' distance implies that inference can be done through the mean and covariance matrix - ...
6
How to show this curious combination of Exponential order statistics has a Chi-squared distribution?
Upon reparameterizing and rescaling ($\chi^2$ distributions are special Gamma distributions), the question is equivalent to showing that
$$Z_r = \sum_{i=1}^r Y_i + (n-r)Y_r$$
has a Gamma$(r)$ distribution. Let's rewrite this suggestively as
$$Z_r = nY_1 + (n-1)(Y_2-Y_1) + \cdots + (n-r+1)(Y_r - Y_{r-1}).$$
Exploit these basic (and easily proven) ...
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