# Tag Info

10

No. We might have $P(X=n)=2^{-n}$. Then $\Bbb E[X]=2$ and $\Bbb E[e^X]=\sum_n 2^{-n}e^n=\infty$

8

It might be easier to think about if we abstract a little. Brownian motion is just a continuous, time dependent random variable determined by some probability space $(X,\mathcal F, P)$. Let us view it as a function of two variables, $f(x,t)$. If $x$ is fixed, we have a continuous function, $f_x(t)=f(x,t)$, and we can compute it's integral $F(x,t)=F_x(t)=\... 6 Let$X$denote any value between$1-5$, then the optional sequences are:$XXX66X6X666XX66$Calculate the probability of each sequence:$P(XXX66)=\frac56\cdot\frac56\cdot\frac56\cdot\frac16\cdot\frac16=\frac{5^3}{6^5}P(X6X66)=\frac56\cdot\frac16\cdot\frac56\cdot\frac16\cdot\frac16=\frac{5^2}{6^5}P(6XX66)=\frac16\cdot\frac56\cdot\frac56\cdot\...

6

The moment generating function of a Poisson random variable $X$ with parameter $\lambda$ is $$\mathbb{E}\left[e^{tX}\right]=\sum_{n=0}^\infty \frac{e^{-\lambda}e^{nt}\lambda^n}{n!}=e^{\lambda(e^t-1)}$$ which converges for all $t$, and in fact is the restriction to $\mathbb{R}$ of an entire function. In particular, the power series for the moment generating ...

5

Of course you don't need to learn statistical physics to do probability theory and there are many areas of probability that are unrelated to physics. However, it certainly wouldn't hurt to at least learn some of the basics. For a very mathematical approach to the subject, I highly recommend the ongoing book project Statistical Mechanics of Lattice Systems: a ...

5

To be clear, you want to find $P_n(t)$ for a birth-death process with constant birth rate $q(i,i+1)=\lambda$ and constant death rate $q(i,i-1)=\mu$? This particular birth and death process is exactly the $M/M/1$ queue. As you can see, under "Transient solution", there is a solution for the probability mass function dependent on time for a particular state.

5

We have to assume that the underlying probability space is complete; otherwise the assertion might fail. So, suppose that $(\Omega,\mathcal{A},\mathbb{P})$ is a complete probability space and $(X_t)_{t \in [0,1]}$ a process with almost surely continuous sample paths, i.e. there exists a null set $N \in \mathcal{A}$ such that $$[0,1] \ni t \mapsto X_t(\omega)... 5$$E[z_1]=\frac{E\Big[{\sqrt{a_1^2-4a_0}}\Big]}2\\=\frac18\left(\int_{-1}^1\int_{-1}^1\sqrt{x^2-4y}dydx\right)\\=\frac1{48}\left(\int_{-1}^1(x^2+4)^{\frac32}-(x^2-4)^{\frac32}dx\right)\\=\frac1{24}\left(\int_0^1(4+x^2)^\frac32dx+i\int_0^1(4-x^2)^\frac32dx\right)\\=\frac{1}{24}\left(\left(\frac{11\sqrt5}4+6\sinh^{-1}\frac12\right)+i\left(\frac{9\sqrt3}4+\pi\...

5

The other solutions work perfectly for a modest number of rolls. If you were interested in a larger number, you might find a recursion helpful. Let $P(n)$ be the probability that your game ends in exactly $n$ rolls. Thus your problem is asking for $P(5)$. We note that $P(1)=0,\;P(2)=\frac 1{6^2}$. For $n>2$ we remark that the first roll is either a $6$...

4

These imply $E[X]=c$ and $E[Y]=d$. However, a constant conditional expectation does not imply independence. Consider any independent random variable $(X,Z)$, and by defining $Y=ZX$ create $(X,Y)$ for which $E[ZX|X]=XE[Z|X]=XE[Z]=d$.$X$ and $Y=ZX$ are not necessarily independent.

4

We see that $$P(Y|X) \cdot P(X) = \frac{P(Y,X)}{P(X)}\cdot P(X) = P(Y,X) = P(Z),$$ Where $Z$ is defined as the variable over the cartesian product of the supports of $Y$ and $X$ (as in, a realization $z$ of $Z$ is in $S_Y \times S_X$). Hence, 1. is the more appropriate answer.

4

Here is a solution that uses SLLN and the additive property of the Poisson distribution. Let $(N_t)_{t \geq 0}$ be a Poisson process of unit rate. Then by the SLLN, together with the inequality $$\frac{N_{[t]}}{[t]+1} \leq \frac{N_t}{t} \leq \frac{N_{[t]+1}}{[t]},$$ it is easy to check that $N_t / t \to 1$ as $t \to \infty$ a.s. Now let $T_k = \lambda_1 ... 4 Each coupon appears$\log n$times on average. If I treat the number for a single coupon as a binomial, the variance is also asymptotically$\log n$. Now treat each of the counts as a normal variable$\mathcal N(\log n,\log n) $. Expectation of the maximum of gaussian random variables shows the maximum of$n$independent standard normal random ... 4$\sum_iY_i=0$. That can only happen for independent variables if they are all almost surely constant. Thus there are no such variables. 4 Fix a nonnegative integer$n$, an integer$m$with$0\leq m\leq n$, and a nonnegative integer$l$. Let$q(n,m,l)$be the probability that the union of random$m$-subsets$S_r$, for$r\in[l]$with$[l]:=\{1,2,\ldots,l\}$, of the set$[n]:=\{1,2,\ldots,n\}$is precisely$[n]$, where the random sets are chosen uniformly randomly and statistically independently.... 4 Since Brownian Motion has continuous paths with probability 1 thus$\int_{0}^{t} W_s ds$is well define. Now set$f(x)=x^3$. By application of Ito's lemma, we have $$f(W_t)=f(W_0)+\int_{0}^{t}f'(W_s)dW_s+\frac{1}{2}\int_{0}^{t}f''(W_s)ds$$ $$W^3(t)=3\int_{0}^{t}W^2_s\,dW_s+3\int_{0}^{t}W_sds$$ therefore $$\int_{0}^{t}W_sds=\frac{1}{3}W^3(t)-\int_{0}^{t}W^2_s\,... 4 The former parts are all right, except you didn't pay attention to the definition region of t. The joint distribution of t and x is:$$ f(x,t)=f_{0}(t\mid x)f(x)=3x/8~~~1\leq x\leq 2, ~x\leq t \leq 2x $$From the graph you draw, we could see the integration should be:$$P[T>3]=\int_3^4 dt \int_{t/2}^{2}3x/8 \, dx=0.171875$$4 An “in real life” interpretation of the last identity is that the number of different chaired even-sized committees from n people equals the number of chaired odd-sized committees from n people. Note that this is not an identity if n=1, as the left side is 1 and the right side is 0. However, it is an identity for n>1. A combinatorial proof (... 4 So both the fourth and the fifth rolls need to be sixes:$$P(4^{th},5^{th}\mbox{ rolls are sixes}) = \frac{1}{6^2}$$There are following possible combinations of rolls for the game to not end until the fifth roll:$$(XXXOO), (XOXOO),(OXXOO)$$where X,O represent non-six and six, respectively. Thus, the probability is:$$P(\mbox{end in fifth trial}) = \... 4 Here is simplified solution. In order experiment to end at 5th trial, the last two rolls must be 6. So we have, NNN66$Pr=5^3/6^5$6NN66$Pr=5^2/6^5$N6N66$Pr=5^2/6^5(N <6)$Add all probabilities. PS. The last two roll must be (6,6), so we get$\frac {1}{6^2}$The third roll can't be 6, so we get$\frac {5}{6}$And first two rolls can be anything ... 4 Since$\int f \, d\mu_n \to \int_0^1 f(x) \, \lambda(dx)$for all continuous bounded funtions$f$, we know that$\mu_n$converges weakly to$\lambda|_{[0,1]}$. Now it follows directly from the portmanteau theorem that $$\int f \, d\mu_n \to \int_0^1 f(x) \, \lambda(dx)$$ for all bounded measurable functions$f:[0,1] \to \mathbb{R}$with$\lambda(D_f)=0$... 4 No, your process does not make any sense. Even if the mean of$X_k$is$1$, this does not mean that $$\frac {X_1 + \dots + X_N}{N} = 1$$ for some$N$. This is a huge misunderstanding; I'll try to briefly explain why is it so. Random variables are not numbers When talking about a random variable, you're talking about a function. When you write$X_1$, ... 4 Actually, the answer to this lies in quite a bit more advanced topic called rough path theory (beware: PDF). A rough path is a way of "enhancing" a$\alpha$-Hölder continuous path with some extra information. A rough path is an ordered pair,$\textbf{X}=(X, \mathbb{X})$where$X\colon [0,T]\to V$where$V$is some Banach space (typically$\Bbb{R}$) and a ... 4 If$n$is the number of tosses, then$T=n-H$, so$H-T=2H-n$. Since$n$is large,$H$has a close to normal distribution, mean$n/2$, and variance$(n)(1/2)(1/2)$. So$2H-n$has close to normal distribution, mean$0$and variance$n$. Let$W$be a normal with mean$0$and variance$n$. Let us find the mean of$|W|\$. This is \int_{-\infty}^\infty \frac{|w|}... 4 Somewhat more generally, let X be any random variable whose moment generating function M(z) = \mathbb E[e^{tX}] is analytic in a neighbourhood of 0. This says that the series \sum_{j=0}^\infty \mathbb E[X^j] t^j/j!  has positive radius of convergence, i.e. |\mathbb E[X^j]| \le C D^j j! for some constants C, D. Then we have uniqueness in the ... 4 You can make a table. The names of the rows are the outcomes of the first dice (d1). The names of the columns are the outcomes of the second dice. In total we have 36 possible outcomes. And the cells marked with \color{red}x are the outcomes where the second dice has a greater outcome than the first dice.  \begin{array}[ht]{|p{2cm}|||p{0.5cm}|p{0.5cm}|p{... 4 Since \mathcal F is any \sigma-algebra on [0,1], this is essentially the general case of: Halmos [1] showed that the range of a non-negative, finite measure is a closed subset of real numbers. [1] Halmos, Paul R. On the set of values of a finite measure. Bull. Amer. Math. Soc. 53 (1947), no. 2, 138--141. http://projecteuclid.org/euclid.bams/... 3 First, rewrite the exponenent as \begin{align*} \exp \left( n\left( a\frac{S_n}{n} - b \right) \right) \end{align*} By the Law of Large Numbers, we have a S_n/n- b \to a \mathbb{E}(\xi_1) - b a.s.. So a S_n -bn \to - \infty a.s. if and ony if b>a \mathbb{E}(\xi_1) =0, and thus \begin{align*} \exp( a S_n - b n ) \to 0 \text{ iff } b > 0. \end{... 3 Let X denote the random variable counting the number of times heads occurs in 2n independent flips of a fair coin. Via the binomial distribution: Pr(X=k) = \dfrac{\binom{2n}{k}}{2^{2n}} More specifically, Pr(X=n) = \dfrac{\binom{2n}{n}}{2^{2n}} Via stirling's approximation, \binom{2n}{n}\sim \frac{4^n}{\sqrt{\pi n}} as n grows large, so the ... 3 This can be solved using the general solution of the Gaussian integral \int\!d^nx\,\exp\Bigl(-\frac12 x^T A x\Bigr) = \sqrt{\frac{(2\pi)^n}{\det A}}.$$In your case, you we have that$$\mathbb{E}\left[e^{x^{T}Ax}\right] = \sqrt{\frac{1}{(2\pi)^n \det \Sigma}} \int\!d^nx \exp\Bigl(-\frac{1}{2} x^T \Sigma^{-1} x\Bigr) e^{x^T A x}=\sqrt{\frac{1}{(2\pi)^n \...

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