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Since you asked this we have updated the Rayleigh wikipedia entry with the unbiased MLE for the Rayleigh parameter as well as confidence intervals, which are conveniently functions of the $\chi^2$ distribution. In particular: Given a sample of N i.i.d. samples $x_i$ from the Rayleigh distribution with parameter $\sigma$, $\widehat{\sigma^2}\approx ... 0 The technique known as "inverse transform sampling" is useful to know. If$X$is any continuous random variable with cumulative distribution function$F_X$,$F_X(X)$has a Uniform distribution on$[0,1]$. From this, note if you define$F_X^{-1}(x) = \inf\{ c : F_X(x) \geq c \}$where$c \in (0,1)$, then$F_X^{-1}(U)$has the distribution of$X$when$U$... 2 Imagine that each candy has an ID number. There are$\binom{48}{5}$equally likely ways to draw$5$candies. Now we find the probability that Jimmy is unhappy with the bag, that is, that in his sampling he draws$0$or$1$strawberry. There are$\binom{6}{0}\binom{42}{5}$ways to draw$0$strawberry and hence$5$non-strawberry/ There are ... 0 Outline: Suppose that a program has$n$lines. Then the number$X$of errors in the program has Poisson distribution with parameter$\lambda=n(0.015)$. We want$\Pr(X\le 1)\le 0.10$. So we want$e^{-\lambda}(1+\lambda)\le 0.10$. This inequality does not have a nice algebraic solution for$\lambda$. However, a little calculator experimentation will yield a ... 0 You want $$P(g(X) > y) =\int_y^\infty \lambda\exp (-\lambda u) du = \exp (-\lambda y)$$ but this is also, assuming$g$is increasing from$[0,1]$to$\Bbb R^+$: $$P(X > g^{-1}(y)) = 1-g^{-1}(y)$$that is: $$g^{-1}(y) = 1- \exp (-\lambda y) g(y) = -\frac{\log(1-y)}\lambda$$and you check that$g$is increasing from$[0,1]$to$\Bbb R^+$. 0 The Central Limit Theorem examines sums of random variables from which we subtract the mean of the sum and then divide the whole by the standard deviation of the sum: let$Y_1,...,Y_n$be random variables and define$S_n \equiv \sum_{i=1}^nY_i$. Then the CLT examines the random variable $$Z_n = \frac {S_n - E[S_n]}{\sqrt {\text {Var}(S_n)}}$$ and what is ... 1 Let$p(x, y) = A(x + y)^2$. We know that since$p(x, y)$is a probability density function, we must have$\int_0^1\int_0^1p(x, y) \mathrm{d}x \mathrm{d}y = 1$. You need to solve this integral for$A$.$p(y | x) := \frac{p(x, y)}{p(x)}$. To calculate this, you need to find the marginal distribution of$x$, which you get by integrating$y$out of the ... 1 The lack of memory of the exponential distribution can be used to produce conceptual proofs that, for every$n\geqslant2$,$G$is distributed as the maximum of$(n-1)$i.i.d. random variables each exponentially distributed with parameter$a$. Since, however, the OP failed to explain their background, here is a direct, hands-on, approach. Consider ... 1 I suggest adapting this code I've taken from the link I gave in my first comment. >> A = rand(700,1); >> MAX = max(A); >> STD = std(A); >> MAX = max(A); >> MIN = min(A); >> STEP = (MAX - MIN) / 1000; >> PDF = normpdf(MIN:STEP:MAX, M, S); >> plot(MIN:STEP:MAX, PDF); In your case the distribution isn't random, ... 1 Try hist(data) to get a histogram. You can also try [f,xi] = ksdensity(data); figure() plot(xi,f); to get a smoothed histogram. -1 There is a Markova property as$\min A_i\sim exp$And it is correct with your step as$\max A \le min\implies\max=min$. I think the difficulty is how to get the distribution function from the wrok you have done 1 It's called a Poisson binomial distribution. You can find useful information about it on Wikipedia. 1 The total number of trials is indeed$r+x$. There are several closely related but not identical definitions of the negative binomial. Definition 1: The random variable$T$is the number of trials until and including the$r$-th success. Definition 2: The random variable$U$is the number of trials up to but not including the$r$-th success. Definition 3: ... 1 The central limit theorem is offtopic to describe the convergence in probability of$X_n$. Note instead that$X_n$is the mean of the i.i.d. random variables$Y_k^2$hence, by the law of large numbers,$X_n$converges to$E(Y_1^2)=1$(in probability, using the weak law of large numbers, and in fact almost surely, using the strong law of large numbers). The ... 0 Firstly, since$Y_i \sim N(0,1)$then$X_i:=Y_i^2 \sim \chi^2_1$for all$i \in \mathbb N$. Quoting Wikipedia: The sample mean of$n$i.i.d. chi-squared variables of degree$1$is distributed according to a gamma distribution with shape$α$and scale$θparameters: $$\bar X_n = \frac{1}{n} \sum_{i=1}^{n} X_i \sim \mathrm{Gamma}\left(\alpha=n/2, \theta= ... 0 Given an alphabet of k characters, the probability of there being any match between any character in S1 and any character in S2 given equal probability of each character, is$$ 1 - (\frac{1}{k} \frac{1}{k-1} ... \frac{1}{k-m})^n $$1 If you mean the corresponding character should be equal: the alphabet contains of k characters, there is for each position a probability of 1/k of a match. The probability that there is no match equals (1-\frac{1}{k})^{\min(m,n)}, hence the probability of at least one match is 1-(1-\frac{1}{k})^{\min(m,n)}. 0 Suppose X_1,\ldots,X_N are independent and identically distributed normal RVs with mean \mu and std dev \sigma. The LHS is P(\frac{X_1+\cdots+X_N}{N} \geq \theta) (why?), and the RHS is P(X_1 \geq \theta)^N. The event X_1 \geq \theta,\,\ldots,\,X_N \geq \theta implies \frac{X_1+\cdots+X_N}{N} \geq \theta, and therefore, ... 1 Edit Based on OP Comments Actually, I realised that the sum of two bernoulli rvs with different ps will result in *under*dispersion, so perhaps underdispersed? If you don't want to be associated with dispersion models, then why not "Heterogeneous Binomial Sum", its clearer than generalized binomial, as there are several ways you could generalize it. 0 It´s all good. However for calculating the variance I do suggest using the following identity, which is easily derived(\mu_x is E(X), in case it is not clear):$$\begin{align} var(X) &= E\left((X-\mu_x)^2\right)\\ &= E\left(X^2 - 2X\mu_x + \mu_x^2\right)\\ &= E(X^2) - E(2X\mu_x) + E(\mu_x^2)\\ &= E(X^2) - 2\mu_xE(X) + \mu_x^2\\ &= ... 0 1) If you get a negative constant which multiplied by a non-negative function is supposed to give a density, you can be certain you did something wrong without asking. Such reasonableness checks are elementary to this kind of exercise. 2) The median will have at least half the probability is to the left of it (including itself) and half to the right ... 1 One is for discrete variables, the other for continuous. They are also interpreted differently. The pdf is a probability "density". If f(x) is the pdf, f(x) doesn't tell you the probability of getting x (In fact, the probability of getting precisely x is 0). The way you use pdf's is through integration. So, the probability of getting a value that is in a dx ... 1 To be precise, a probability mass is a probability distribution. However, it is not a probability density - probability masses are discrete, while probability densities are continuous. 0 So there are2^5$possible sequences in terms of break or not break. 16 of those 32 have 3 or more breaks. More precisely, there are 10 ways to get 3 breaks, 5 ways to get 4 breaks, and 1 way to get 5 breaks. However, these sequences are weighted. The easiest is the 1 way to get 5 breaks, this probability is 0.05^5. The second easiest is 5 ways to get ... 0 The interarrival times have exponential distribution mean$\frac{1}{\lambda}$, so yes. 0 You can use the coefficient of variation (CV) to compare the homogeneity/ variability between the two distributions. The CV is given by $$CV=\frac{s}{\bar{x}}$$ where$s$denotes the standard deviation of the sample and$\bar{x}$the mean of the sample. The$CV$is measured in percent units, which means that it is a number (free of specific measurement ... 0 For different distrbutions, mean and standard deviation control different features of the pdf. For the Normal distribution: the larger the mean, the more to the right the spike will be, and the larger the variance, the more steeply it will come down from the spike (i.e., the spike will be less spread out horizontally). EDIT For the Beta distribution ... 0 Answer: Tossing an unfair coin N times with a probability of success being r and you get$N_{1}$times heads and$N_{2}$times tail follows a beta distribution and the pmf is given by $$\dfrac{r^{N_{1}}(1-r)^{N_{2}}}{\beta(N_{1}+1,N_{2}+1)}$$$N_{1}$and$N_{2}$define the shape parameters of the beta distribution. It is r that decides the peak. r ... 0 The binomial distribution can be used to determine the probability that exactly$k$out of$n$independent and identical trials will succeed. If the probability that a trial succeeds actually depends on how many trials you've already carried out, then the trials are not identical. If, in addition, the probability of success depends on how many of the ... 0 I don't see the problem. Your success probability depends on$n$, and decreases as$n$increases. In this case, your success probability is converging to zero just fast enough that the probability of observing exactly one success is roughly constant. Note that you should be very careful about summing over$n$, since the success probability is different ... 0 No, whether or not the parameters are equal has nothing to do with whether or not they are independent. They can have the same parameter value and be dependent or independent and likewise if their parameter values differ. The only ways to figure out whether or not they are independent is to check whether or not their joint distribution function (cdf) ... 2 Suppose$W$and$X$are independent Poisson-distributed random variables each with expected value$1$. Let$Y=W+X$. Then$Y$is Poisson-distributed with expected value$2$. So$X$and$Y$are Poisson-distributed random variabes with different expected values, but they are certainly not independent. 2 Hint: Let$X$be the number of Poisson events in the first hour, and$Y$the number of Poisson events in the first two hours. If$X$has parameter$\lambda$, then$Y$has parameter$2\lambda$. The two random variables are not independent. 1 The question, in my opinion, extends to the debate "Bayesian vs classical (or frequentist) statistics". According to Wikipedia there are two major differences in the frequentist and Bayesian approaches to inference: In a frequentist approach to inference, unknown parameters are often, but not always, treated as having fixed but unknown values that are not ... 2 If$(X_i)$is i.i.d.,$(Y_i)$is i.i.d. and$(X_i)$and$(Y_i)$are independent, this follows from the central limit theorem applied to the i.i.d. sequence$(Z_i)$defined by $$Z_i=X_i-Y_i-E(X_1)+E(Y_1).$$ To wit, considering the events $$A_N=[S_2-S_1\geqslant E(S_2)-E(S_1)],$$ one gets $$... 2 Firstly solve (b) so that you will know the value of a at the time that you will plot the function. f is a joint valid pmf if (i) f(k,l)\ge0 and (ii) \sum f(k,l)=1. So, in order that all the values of f add up to 1 (already mentioned in the comments), we solve the equation$$1=a(1^1+1^2+1^3+2^1+2^2+2^3+3^1+3^2+3^3)=56a$$which implies that ... 0 You need a continuous probability distribution since waiting time is continuous. Since waiting time Y is said to be proportional to the given wait time, the probability distribution function f_Y(y) of Y should be$$f_Y(y)=\begin{cases}cy, & 1\le y\le13\\ \\ 0,& \text{elsewhere}\end{cases}$$where 0<c. Since the integral of f has to be ... 1 Hint: \lceil X\rceil\geq k\iff X>k-1 so:$$P\left\{ \lceil X\rceil\geq k\right\} =P\left\{ X>k-1\right\} $$1 Since the probability of two wait times are proportional, that implies that there is a non-zero probability of Y being a given number, hence the distribution should be discrete (i.e., a pmf). Since Y is the "approximate" wait time, lets break up the interval into 13, 1 minute increments, with approximate values on the integers between 1 and 13 (since times ... 1$$E \left[ f\left(\frac X {|X|}\right) g(|X|) \right]= \frac 1 {Z^n}\int f \left(\frac x {|x|}\right) g(|x|) \exp \left(-\frac 12 |x|^2\right) dx_1\cdots dx_n $$now note x = r x', with |x'| = 1 and r>0, and M(r,n) beeing the size of the sphere of \mathbb R ^n of radius r yields$$E \left[ f\left(\frac X {|X|}\right) g(|X|) \right]= \frac ... 0 You are right about$X_1^2+\cdots+X_n^2$, and there's really nothing more you need to say about it, unless the definition of the$\chi^2_n$distribution that you're working with is something other than the distribution of the sum of squares of independent standard normally distributed random variables. For the distribution of$X/\|X\|$, I think I might ... 0 I took an easier approach and I think I got it. Just to be a bit more general I'll be using$(\mu_1, \mu_2 ) $as the means of$(\epsilon_1, \epsilon_2) $even though the question stated they were zero. Write the standard normal PDF and CDF by$\phi$and$\Phi$. Define $$Pr(\epsilon_2 > \alpha) = 1 - F_2 (\alpha)$$ and multiply and divide by this term: ... 1$(z+3)^2$is a noncentral$\chi_1^2$, so$(z+3)^2-8$is a noncentral$\chi_1^2$with a location and scale shifts. 0 We want to measure the expected “deviation” of a random variable from it's expected value. What comes first in mind is the expected difference between the r.v. and it's expected value, that is$E\bigl(X-E(X)\bigr)$. But this number is always zero. -- Next, as drhab pointed out, is the average distance, i.e., the absolute value of the difference, between ... 1 You can use the normal approximation (indeed this is due to the Central Limit Theorem) to the Poisson distribution (see here). You have that $$X \sim Poisson(\lambda=36)$$ (where$\lambda>20$as required in the link), therefore $$X \sim N(\mu=36, \sigma^2=36)$$ approximately, which can be equivalently formulated as ... 1 If I want to 'measure' how much random variable$X$is expected to differ from its expected value than intuitively I think of things like$\mathbb E(|X-\mathbb E(X)|)$or$\mathbb E(X-\mathbb E(X))^2$. Another possibility in the same line is the root of the variance, named deviation. Have a look at the anwer of Stefanos when it concerns second generating ... 1 The idea one can say, is that the variance is a measure of how the values of a random variables are spread in the space. So if you have a random variable that takes many values identical to the mean value and few values far away from the mean then that means that distribution is more focused in a smaller space, hence the variance is smaller. In other words ... 0 $$\operatorname E(XZ)=\operatorname E(X^2+XY)=\operatorname EX^2+\operatorname E(XY)$$ and, since$\operatorname{Var}X=\operatorname EX^2-(\operatorname EX)^2=\lambda$, we have that $$\operatorname E(XZ)=\lambda+3\lambda^2.$$ Finally, $$\operatorname{Cov}(X,Z)=E(XZ)-\operatorname EX\operatorname EZ=\lambda+3\lambda^2-3\lambda^2=\lambda.$$ Edit. As ... 0$X$is the random value,$x$is a value$X$can take. Formally,$X$is a function of an alea: $$X:\Omega\to A$$ and$x$is one element of$a$. For example,$X$can be tomorrow's temperature, and$x = 20$degrees. And for the second point, it is true:$f_{X,Y}(x,y)$is the probability that the$(X,Y)$equals$(x,y)\$.