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## Hot answers tagged probability-distributions

3

Try $X \sim N(0,1)$ and $Y=X$ when $|X|\lt k$ and $Y=-X$ when $|X|\ge k$ for some non-negative $k$. Then $X$ and $Y$ have standard normal distributions and $cov(X,Y)$ is a continuous increasing function of $k$, negative when $k$ is close to $0$ and positive when $k$ is large. So for some $k$ you will have $cov(X,Y)=0$

3

Note that this is a "small"problem. Since everything is a multiple of $25$, we might as well assume that we start with $4$ gold coins, and we bet until we either have $6$ gold coins or none. Let random variable $X$ be the number of bets until the game is over. It looks as if we are only being asked about $E(X)$. Let $e_1$ be the expected further length ...

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I think you mean the bernoulli distribution otherwise the binomial distribution is $\binom{n}{x}p^x(1-p)^{n-x}$ The idea though is the same. In a bernoulli distribution you have a random variable that takes the values $1$ if you have success (you can think that it is head at a coin) and $0$ if you fail (tails at a coin). When you have succes then the ...

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Durrett's probability book appears to still be free (on author's page). Your subject is embedded in Chapter 3 Central Limit Theorems (Weak Convergence, Characteristic Functions etc., leading to Continuity Theorem 3.3.6). Rick Durrett's Probability: Theory and Examples book Theorem 3.3.6. Levy's continuity theorem: Let $\mu_n$, $1\leq n \leq \infty$ be ...

2

We want $$\frac{\Pr((X\gt 0) \cap (X+Y\gt 0))}{\Pr(X+Y\gt 0)}.$$ By symmetry the denominator is $\frac{1}{2}$. For the numerator, we want the probability that $(X,Y)$ lands in the part of the plane that is to the right of the $y$-axis, and above the line $y=-x$. By symmetry this is $\frac{3}{8}$.

2

Let $X=D^{-c}H$ with obvious notations and $x\gt0$, then $[X\gt x]=[H\gt xD^c]$ hence, by independence of $H$ and $D$, $$P(X\gt x)=E(\exp(-\lambda xD^c)).$$ Since the distribution of $D$ is known, at least in principle this expectation can be computed. If $D$ is uniform on $(0,1)$, the change of variable $v=tu^c$ yields $$... 2 If (X_i) is i.i.d., (Y_i) is i.i.d. and (X_i) and (Y_i) are independent, this follows from the central limit theorem applied to the i.i.d. sequence (Z_i) defined by$$ 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

Hint: Start with the cdf of $W:=-2\ln(Y^4)$, for $0<y$. You have that \begin{align*}F_W(y)&=P(W\le y)=P(-2\ln(Y^4)\le y)=P(-8\ln Y \le y)=P(\ln Y \ge \frac{y}{-8})=\\&=P(Y\ge e^{-\frac{y}{8}})=1-P(Y\le e^{-\frac{y}{8}})=\\&=1-F_Y(e^{-\frac{y}{8}})\end{align*} Taking the derivative: ...

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When you know evey moment of a bounded random variable you know its law: a random variable is caracterized by its Fourier transform $E\exp iu X$. In the case of a bounded $X$, you have $$E\exp iu X = E\sum_{n=0}^\infty \frac{(iu X)^n}{n!} =\sum_{n=0}^\infty EX^n\frac{(iu )^n}{n!}$$ You can swap $E$ and $\sum$ thanks to the Fubini theorem, because $X$ is ...

2

(Assuming you have 8 identical balls and 6 boxes): Before checking the below answer check the link I gave in the comments above. Also this is the same thing I tried to point in your last question:Two probability questions. A:= B1 is not empty. B:= B1 and B2 are not both empty. Size of sample space = $\binom{13}{5}$ P(B1\ is\ empty) = ... 2 Following the analysis here: \begin{align} F(x) &= \int_0^{\infty} \frac{dy}{y} e^{-y-\frac{x}{y}}\end{align} Subu=y+\frac{x}{y}, then $$y = \frac12 \left (u \pm \sqrt{u^2-4 x}\right )$$ $$dy = \frac12 \left ( 1 \pm \frac{u}{\sqrt{u^2-4 x}} \right ) du$$ Then \begin{align}F(x) &= \frac1{4 x} \int_{\infty}^{2 \sqrt{x}} du \left ( 1 - ... 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. 1 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 equation1=a(1^1+1^2+1^3+2^1+2^2+2^3+3^1+3^2+3^3)=56a$$which implies that ... 1 \hat\lambda= \frac{n}{\sum_{i=1}^n x_i} to be consistent estimator of \lambda it should be Asymptotically Unbiased, and it's variance goes to zero. Using E\left\{ x\right\}=\frac{1}{\lambda} and E\left\{ x^2\right\}=\frac{2}{\lambda^2} and the fact that x_i are iid, we have Condition 1: \lim_{n\rightarrow \infty} E\{\hat\lambda - ... 1 Failure rate, or hazard rate, or force of mortality of a device/object/life at some time t is best interpreted as the instantaneous exposure to failure/death given survival to time t. Formally,$$h(t) = \frac{f(t)}{S(t)} = \lim_{\Delta t \to 0} \frac{\Pr[t < T \le t + \Delta t \mid T > t]}{\Delta t},$$where T is the future lifetime random ... 1 Try the fact that$$\int_0^\infty \int_0^\infty e^{(-x-y)}dxdy=\int_0^\infty\left(\int_{-\infty}^{-y} e^{u}(du)\right)dy=\int_0^\infty e^{-y}dy=1$$here letting u=-x-y and du=-dx (why?). But I am not sure if this is what you need to find for your problem in general, as prior commenters have pointed out. 1 For the marginal pdfs, note that you split up e^{-(x+y)} into the product of a function of x and a function of y. Or rather, note that it was possible to do this. That says something about whether or not the two variables are independent. What does it say? What then can we say about the marginal pdfs? As for the joint cdf, I think there's some sort ... 1 Define the Bernoulli random variable X_i to be one if the number i is chosen and zero otherwise. Then the number of distinct numbers chosen is X:= X_1+\cdots+X_N. Each X_i is zero with probability \left(1-\frac 1 M\right)^N. Hence, E[X_i]= 1-\left(1-\frac 1 M\right)^N, and by the linearity of expectation,$$E[X] = N\cdot\left(1-\left(1-\frac 1 ... 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 notex = 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 $$E \left[ f\left(\frac X {|X|}\right) g(|X|) \right]= \frac ... 1 One is for discrete variables, the other for continuous. They are also interpreted differently. The pdf is a probability "density". If f(x) is the pdf, f(x) doesn't tell you the probability of getting x (In fact, the probability of getting precisely x is 0). The way you use pdf's is through integration. So, the probability of getting a value that is in a dx ... 1 Hint: You need to apply the method of bivariate transformation of random variables. You can define$$X \sim U(a,b)$$and$$Y\sim \exp(\lambda)$$(to simplify notation, X stands for your d_{mn} and Y for h_{mn}) and then define$$U=X,\qquad V=\sqrt{(1/X)^a}\cdot Y$$Solving for X and Y you find that$$X=U, \qquad Y=\frac{V}{\sqrt{(1/U)^a}}$$from ... 1 They shouldn't be independent: we know that \lvert U_2\rvert<U_1 with probability 1. So, knowledge of the value of U_1 gives us information about the possible values of U_2. With this in mind, it isn't hard to come up with events which disprove this. For instance: P(U_2>\frac{1}{2})>0, P(U_1<\frac{1}{2})>0, but ... 1 The idea one can say, is that the variance is a measure of how the values of a random variables are spread in the space. So if you have a random variable that takes many values identical to the mean value and few values far away from the mean then that means that distribution is more focused in a smaller space, hence the variance is smaller. In other words ... 1 If I want to 'measure' how much random variable X is expected to differ from its expected value than intuitively I think of things like \mathbb E(|X-\mathbb E(X)|) or \mathbb E(X-\mathbb E(X))^2. Another possibility in the same line is the root of the variance, named deviation. Have a look at the anwer of Stefanos when it concerns second generating ... 1 My answer does not give you that distribution but reaches you the possibility to find it yourself. I presume that A, B and C are independent. This should be mentioned in your question. Note that P\left[A<B<C\right]=\frac{1}{6}. If I understand you correctly then you are searching for:$$P\left[B-C\leq x\mid A<B<C\right]=6P\left[B-C\leq ... 1 Given: random variables$A$,$B$and$C$have joint pdf$f(a,b,c)$: Let$Z = B-C$. Then, the cdf of$Z$is$P(Z<z) = P(B-C<z)$: Take the derivative of the cdf wrt$z$to yield the desired pdf, say$g(z)$: $$g(z) = 3(1+z)^2 \qquad \text{ for } -1<z<0$$ All done. Notes The Prob function used above is from the mathStatica package for ... 1 Based on your experiment, I am assuming that you are using the Unpaired Student-t test. Your answer for the t-value is correct, and the t-value can be negative - it does not have to be positive, as you have incorrectly assumed. If the t-value is negative, it means that the mean of the second dataset is greater (i.e. your Lab 2 results are greater than your ... 1 The meaning of the CDF at a point$x_0$is that it answers the following question: What percentage of my variable is less or equal than$x_0$. You know for example that$60$of All users has$10^1$in-degree or less. You also know that almost$100%$of All users has$10^3$in-degree or less. Similarly you also know that$70%$(circa) of Early adopters has ... 1 Let random variables$X_1,X_2,\dots,X_9$be the individual masses. Let$Y=X_1+X_2+\cdots+X_9$be the sum of the masses. If we assume that the$X_i$are independent, then$Y$has (approximately) normal distribution, mean$90$, standard deviation$\sqrt{9}\cdot 5$. We want the probability that a normally distributed random variable with mean$90\$ and ...

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