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## Hot answers tagged probability-distributions

3

The moment generating function of a random variable $X$ is defined by $$M_X(t) = E(e^{tX}) = \begin{cases} \sum_i e^{tx_i}p_X(x_i), & \text{(discrete case)} \\ \\ \int_{-\infty}^{\infty} e^{tx}f_X(x)dx, & \text{(continuous case)} \end{cases}$$ If we express $e^{tX}$ formally and take expectation $$M_X(t) = E(e^{tX}) = 1 + tE(X) + ... 3 Let X \sim \operatorname{Normal}(\mu,\sigma). Then the MGF of X is given by$$M_X(t) = e^{\mu t + (\sigma t)^2/2}.$$The MGF of the sum$$n\bar X = \sum_{i=1}^n X_i$$where each X_i is IID as X, is simply$$M_{n \bar X}(t) = M_X(t)^n = e^{n (\mu t + (\sigma t)^2/2)}.$$This demonstrates that n \bar X is also normal, but with mean \mu^* = n ... 2 We are told that X = X_1+\dotsb+X_N, where N\sim\text{Pois}(\lambda). a) Essentially, you are asked to compute or give P(X = k|N = n). If I tell you that N = n, then the sum of n independent Bernoulli trials follows a binomial distribution with (n,p). Hence, X|N \sim \text{Bin}(n,p). b) I believe that you are essentially being asked to ... 2 Write I_n = E[X^n]. You have$$ I_n = \int_{-\infty}^{\infty} x^n \cdot f_X(x) \, dx $$where f_X is the probability density function of X. In your case,$$ I_n = \int_0^{\infty} x^n \cdot (\lambda e^{-\lambda x}) \, dx. $$Applying integration by parts with u = x^n and dv = \lambda e^{- \lambda x} \, dx (so v = -e^{- \lambda x}), you have ... 2 Given information: Pr(X=x) = .8\cdot .2^{x-1} for each x\geq 1 Side note: it is worth checking on your own that this is indeed a probability distribution. I.e. that \left(\sum\limits_{n=1}^\infty Pr(X=n)\right) = 1. It is and it does. (a) Find the probability that a point requires at least two shots. I.e. find Pr(X\geq 2). By ... 2 You can solve it by inspection. Just fill in the table:$$\begin{array}{l:l}x & \dfrac{2^x~\mathsf e^{-2}}{x!} & \displaystyle\sum\limits_{k=1}^x\dfrac{2^k~\mathsf e^{-2}}{k!} \\ \hdashline 0 \\ 1 \\ 2 \\ 3 \\ 4\\ 5 \\ \hline 6 \\ \vdots \end{array}Stop when the third column exceeds 0.99. Do you have access to a spreadsheet ... 2 There are two schemes to define the the integral \int_{\Omega}X(\omega)dP(\omega), for X a random variable taking values in a Banach space --- Bochner integral and Pettis integral --- depending on the strong or weak measurability of X respectively. Click here for Bochner integral, and here for Pettis integral, both from Wikipedia. 2 continue from your last line that \sum_{n=0}^{\frac{k-b}{a}} e^{\lambda} \frac{\lambda^n}{n!} - \sum_{n=0}^{\frac{k-b-1}{a}} e^{\lambda} \frac{\lambda^n}{n!} = e^{\lambda}\frac{\lambda^{\frac{k-b}{a}}}{\frac{k-b}{a}!} Since k is an integer calculated from k = aq + b (for some q satisfying X-distribution) 2 Here is a guideline. Some details are left as smaller exercises, but I really encourage you to fill them up. I'll first recall how to prove that a sequence of random variables converges almost surely, before adapting the argument to the limsups. Convergence of sequences of random variables Let (a_n)_{n \in \mathbb{N}} be a sequence of real numbers, and ... 1 \begin{align} &\Pr[X_1 < X_2 < X_3] \\ =& \sum_{x_2}\Pr[X_1 < x_2\ \land\ x_2 < X_3\mid X_2 = x_2] \cdot \Pr[X_2 = x_2] \\ =& \sum_{x_2} \Pr[X_1 < x_2] \cdot \Pr[X_3 > x_2] \cdot \Pr[X_2 = x_2] \end{align} assuming X_2 is discrete. 1 In probability one usually works with events. In essence the probability operator P is defined for sets, i.e. in your case we have the set A = \{ 4 < X < 11\} and the set B = \{ 14 < X< 21\}. How the probability operator works is it takes in a set for exampleP( 4 < X < 11 ) = P(A) = P( \{ 4 < X < 11\}),$$which is all ... 1 There are indeed distributions - but no true functions - which can be interpreted as having infinite probability density at a point. The classical example is the Dirac delta function (which is not really a function). 1 No, you're ignoring various dependencies. In particular, where you write "These two would be independent", they're not; you're including cases where Y_1^{(n-m)}\lt Y_2^{(m)}, despite previously assuming that Y_1^{(n)} is greater than the first m values. The probability that exactly k of the values are in (x-a,x] and the other n-k are at most ... 1 First, randi([1 2]) will give you a uniform distribution from 1 to 2. And you said that you will count the difference between number of heads and number of tails. In general, if the coin-flip experiment is fair, the probability of getting a head should be the same as the probability of getting a tail = 0.5. That means the difference should be zero. But in ... 1 It is exactly what you have to do. By definition, to be a density function, f has to be positive and its integral has to be 1. If I am not mistaken, you should find a=6/5. 1 Here's why what you did didn't work. Let X,Y be two independent normally distributed random variables with mean 2 and variance 1 (this is what you sampled from with your R code). This means that X+Y is normally distributed with mean 4 and variance 2, which implies that$$ \frac{X+Y - 4}{\sqrt{2}}$$is a standard normal random variable, so it follows that ... 1 You will have to sum up a number of mutually exclusive probabilities: The last winning match has to be B, and B must win any 2 of the preceding ones, hence \binom22 + \binom32 + \binom42 cases BBB,\;ABBB,\; BABB,\; BBAB,\; AABBB,\; ABABB,\; ...... with probabilities 0.3*0.8*0.3 + 0.7*0.8*0.3*0.8 + ..... 1 Split it into disjoint events, and add up their probabilities: P(AABBB)=0.7\cdot0.2\cdot0.3\cdot0.8\cdot0.8 P(ABABB)=0.7\cdot0.8\cdot0.7\cdot0.8\cdot0.8 P(ABBAB)=0.7\cdot0.8\cdot0.3\cdot0.2\cdot0.8 P(ABBB )=0.7\cdot0.8\cdot0.3\cdot0.8  P(BAABB)=0.3\cdot0.2\cdot0.7\cdot0.8\cdot0.8 P(BABAB)=0.3\cdot0.2\cdot0.3\cdot0.2\cdot0.8 P(BABB ... 1 a) I would probably model the answer as: Pr(x≥2)= 1 - Pr(x=1) as x≥1 1 Your derivation is correct because U=F(X) has uniform distribution:$$ P(U\le u)=P(F(X)\le u)=P(X\le F^{-1}(u))=F(F^{-1}(u))=u,\mbox{ for }u\in[0,1]. $$It is also well known that -\ln(U) has Exponential distribution. 1 1) The conditional distribution of the 8 clients along the interval [0,4] is Uniform, so the conditional distribution of the number of clients in a sub interval (2,4] is Binomial with p=\frac{2}{4} and n=8. Denote it by X, the event of interest is \{X=4\}, so P(X=4) = \binom{8}{4}\frac{1}{2^8}. The answers to 2 and 3 look OK to me. 1 1) Knowing that in the first four hours 8 persons got into the bank, calculate the probability of exactly half of them having entered past the first two hours. I define the random variable Y_i =number of clients in the i-th hour of the opening hours Y_i=number of clients in the i-th hour of the opening hours for i=1,2,3,4 , since ... 1 For a concrete example of how the distributions of the d_{A,i} and d_{i,B} don't determine the distribution of d_{A,B}, consider the following process. We will put A at the origin in the plane, and B at [d_{A,B},0] where d_{A,B} is anything from 3 to 4. Then choose s_i and t_i iid with a uniform distribution (or whatever other ... 1 As$$f_X(1+\delta) = f_X(1-\delta) for $\delta>0$ the mean and the median are the same. Therefore $\mathbb{E}[X]=1$.

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You will have two symmetrical traingles mirrored at x= 1. The mean of this pdf in terms of integration is $E(k) = \int_{0}^{1} k(1-k)dk + \int_{1}^{2}k(k-1)dk = 1$ Just what the other answers have indicated the Mean is simply 1 by symmetry. The mean for the first triangle is the k which splits the area of the triangle into two halves. That \$k_1 = ...

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