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Probability of sum and conditional expectation

Note $(X,Y) \sim \mathrm{Pois}(\lambda)\otimes \mathrm{Pois}(\mu)$. Using generating functions (probability generating functions, moment generating function, or characteristic function all work here), ...
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A negative Kullback-Leibler divergence between two Laplace distributions?

The formula you got is correct for $\log$ in base $e$ (natural log). Sadly, Desmos understands that $\log = \log_{10}$. Replace that by $\ln$ and you'll get the right graph. Besides (using natural log)...
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Expected Maximum Value of 10 Randomly Selected Balls from an Urn

There is a clever method using linearity of expectation. I got this argument from two of joriki's answers: one about cards in a deck and one about the continuous version of points on a line. First, ...
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Accepted

Reconciling Definitions of Conditional Expectation

In the second definition take $\mathcal G=\sigma(Z)=\{Z^{-1}(E): E \,\, \text {Borel in } \mathbb R\}$. If the equation in the first definition holds, you can take $h=1_E$ to get the equation in the ...
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See given that, $$P_n = \sum_{k=2}^n \frac{(-1)^k}{k!} \tag{1}$$ we want to re-write $(1)$ in terms of $P_{n-1}$ and $P_{n-2}$ so, $$P_n = \sum_{k=2}^{n-1}\frac{(-1)^k}{k!} + \frac{(-1)^n}{n!}$$ $... 3 votes Accepted Show$P_n = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots + \frac{(-1)^n}{n!}$given$P_n = \frac{n-1}{n} P_{n-1} + \frac1n P_{n-2}$$$P_n = \frac{n-1}{n} P_{n-1} + \frac1n P_{n-2} \\ P_n - P_{n-1} = -\frac{1}{n}\left({P_{n-1}-P_{n-2}}\right)$$ Let :$V_n=P_n - P_{n-1}Therfore: V_n=-\frac{1}{n}V_{n-1} \implies V_n=(-1)^n\frac{... • 889 0 votes Checking consistency of minimizer given uniform convergence of its objective function If I understand your question correctly, then, a_n^2\to 0, thus O_p(a^2_n)\to 0 in probability. Also, \hat{\theta}_1 \to \theta_0. Therefore \hat{\theta}_2 \to \theta_0. Let me know if this is ... • 2,840 0 votes Dice conditional expectation Found the trick to solve it. Let n be the max value of the die, and (d_1,d_2) be a pair satisfying d_2 > d_1, ie. a favorable event, where the second element of the pair represents your roll, ... • 3,502 1 vote The matching problem in a recursive view The probability that no one sits at the right seat will be that 1 person will sit in the n-1 places that do not belong to them, and the next person will sit in the n-1-1=n-2 that do not belong ... • 280 0 votes Expected Maximum Value of 10 Randomly Selected Balls from an Urn Distribute the 20 numbers in ascending order uniformly on a 0-1 scale, so they're at \frac{k}{21}, for k=1,2,3,...20 On the other hand, in similar vein, the sampled numbers are at \frac1{11},... • 38.7k 1 vote The matching problem in a recursive view Instead of considering full table try to consider emty table. We have two cases. First guest seats on place numbered i, next comes guest numbered i. He can seat on first place or other. If seats ... 2 votes Why Do Fewer Points Result in Larger Variances? There are a lot of unstated assumptions in your question, which is what is leading to your confusion. First of all, what does it mean to say "the variance of these data points?" One ... • 125k 1 vote Why Do Fewer Points Result in Larger Variances? The Crux of this Query is the confusion over "variance" & "variance of variance" ! At the end of the Post , I will re-write what OP has written , to make it Accurate & ... • 6,300 -1 votes Accepted Why is a sequence of random variables not a markov chain? Let's analyze the given information and the problem. The terms Y_1 and Y_2 are i.i.d. Bernoulli(0.5), and for j \geq 3: if \min(Y_{i-1}, Y_{i-2}) = 1, then Y_i is Bernoulli\left(\frac{2}{... 0 votes Where am I going wrong with evaluating this integral? By integration by parts and then L’Hospital Rule \begin{aligned} \int_{200}^{\infty}\left(-x e^{-x / 1000}-1000 e^{-x / 1000}\right) d x = & 1000 \int_{200}^{\infty}(x+1000) d\left(e^{-x / 1000}\... • 16.8k 1 vote Show 1/X_t^2 is bounded where X_t is a random process I'm going to try and provide a very explicit answer to your question. The main trick is to give an upper bound expression that is in general much easier to work with. To start let t \in \mathbb{N} ... • 1,915 1 vote Maximize product of die Your work is wrong (D1 and D2 are not independent) and your script is probably correct. The actual value is \frac{617}{36} \approx 17.13889. To calculate this, consider the expected product for ... • 152k 0 votes Proving that a function is negligible Let me only draw a little picture over here for the case we pick c=2, to help a little bit the intuition. In addition, as John Bentin wrote, let me highlight that we are not obliged to find the ... • 2,734 3 votes Accepted Proof that if Z = X + Y, where X\text{~Bin}(n, p) and Y\text{~Bin}(m, p) and are independent, then Z\text{~Bin}(n + m, p). You may also apply the Law of Total Probability and the independence of X and Y: \begin{align*} \mathbb{P}(Z = z) & = \mathbb{P}(X + Y = z)\\ & = \sum_{x = 0}^{\infty}\mathbb{P}(X = x, Y = ... • 14.4k 1 vote Proof that if Z = X + Y, where X\text{~Bin}(n, p) and Y\text{~Bin}(m, p) and are independent, then Z\text{~Bin}(n + m, p). This is not sufficient; it only shows that the means match. But distributions are much more than their means. What you need to argue is simple: making x binomial(p) decisions "followed by&... • 2,231 0 votes Explain E=\frac{1}{2}\left(E+\frac{2}{3}\right). Either you get to roll (with P = 3/6) or you dont. In case you do, you'll earn E more with P = 3/6 of earning 1 more (rolling 1 or 2 or 3) P = 2/6 of earning 0 (rolling 4 or 5) P = 1/6 of losing 1 (... 2 votes Accepted Given two continuous random variables X and Y with different domains, how can you calculate P(X<Y) given some joint PDF? When you are finding the bounds for integration for computing P(X<Y), you need to take all inequalities into account. We are given that 0\le x\le 20 and 0\le y\le 400. You want to find P(X&... • 70.6k 0 votes Cov(X_t , X_{t-2}) in AR model. Let us work with the AR(1) without the constant term Y_t = X_t-\mu which is more direct. As mentioned by @Henry, you can write \begin{aligned} Y_t &= aY_{t-1} + Z_t \\ &=a(aY_{t-2}+Z_{t-1}... 1 vote Find the probability that at least 2 defective bulbs are drawn, if 4 bulbs are drawn from a box containing 10 bulbs of which 3 are defective. $$\frac{\binom{3}{2}\times \binom{7}{2}+ \binom{3}{3}\times \binom{7}{1}}{\binom{10}{4}}=\frac{1}{3}$$ Number of ways choosing2$defective out of$3$and$2$nondefective out of$7$is$\binom{3}{2}\...
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Im not sure how to use the given inequality here, but one way to shoe the claim is by Jensens inequality. Notice that: $$H(x) = -\sum x_i \log x_i$$ Is concave with respect to $x$. Rewriting your ...