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Hint: let $Y=X/2$; then $\mathbb E\left[Y^n\right]=1/(n+1)$ which is the integral of $y\mapsto y^n$ over an interval I let you find.

4

You had a mistake in your calculation of the moment generating function, when reindexing the sum. In fact, $$1+\sum_{k=1}^{\infty}\frac{(2t)^{k+1}}{2t(k+1)!}=1+\sum_{k=2}^{\infty}\frac{(2t)^{k}}{2tk!}=1+\frac{e^{2t}-2t-1}{2t},$$ which is different what you wrote (you had the sum starting at $0$ and not at $2$). Simplifying this expression yields $$\mathbb ... 2 After the first ball is drawn, seen to be red, and discarded, there will be 21 red balls remaining and 5 non-red balls remaining (what the exact colors of each non-red ball is doesn't matter). Next, since we draw some balls, and then discard them without looking at them, this is effectively the same as though we never drew them at all in the first place ... 2 \mathbb{E}\left[t^{X}\mid\lambda=u\right]=e^{-u\left(1-t\right)} so that \mathbb{E}\left[t^{X}\mid\lambda\right]=e^{-\lambda\left(1-t\right)} and:$$G_{X}\left(t\right)=\mathbb{E}t^{X}=\mathbb{E}\left[\mathbb{E}\left[t^{X}\mid\lambda\right]\right]=\mathbb{E}e^{-\lambda\left(1-t\right)}=\frac{1}{2}\int_{0}^{2}e^{-\lambda\left(1-t\right)}d\lambda=\begin{...

2

The first "=" in (*) fails to be true. Note that $$\left\{ \sup_{0 \leq s \leq t} |W_s|> a \right\} \neq \left\{ \sup_{0 \leq s \leq t} W_s > a \right\} \cup \left\{ \sup_{0 \leq s \leq t} W_s < -a \right\}. \tag{1}$$ Say, for instance, we have a sample path with $\sup_{s \leq t} W_s(\omega)<a$ but $\inf_{s \leq t} W_s(\omega)<-a$, then $\... 2 HINT Perhaps, it helps to understand the problem if you take a look at the following figure: Here, there are$500(a,b)$dots and the corresponding$(X,Y)$dots... The task is (may be) to describe the function below which all the$(X,Y)$dots lie. If the shape of the upper bound of the right hand side dot diagram is the question then we may compute the ... 2 By definition of the exponential distribution, $$\mathbb P(T_1>t)=\mathbb P(T_2>t)=e^{-\lambda t},\qquad t\geq 0.$$ Thus, $$\mathbb P(T_1>kT_2)=\mathbb E\bigl[\mathbb P(T_1>kT_2\mid T_2)\bigr]=\mathbb Ee^{-\lambda k T_2}=\int_0^{\infty}e^{-\lambda k x}\cdot \lambda e^{-\lambda x} dx,$$ where in the last line we are integrating against the ... 1 Write$\langle X Y \rangle - \langle X \rangle \langle Y \rangle = \text{Cov}(X,Y)$, the covariance of$X$and$Y$,$\sigma(X) = \sqrt{\text{Cov}(X,X)}$the standard deviation of$X$, and$\rho(X,Y) = \text{Cov}(X,Y)/(\sigma(X) \sigma(Y))$the Pearson correlation coefficient of$X$and$Y$. Then it can be shown that $$\rho(B,C) \ge \rho(A,B) \rho(A,C) - \... 1 A good thing about (zero-mean) multivariate normal distribution is that its structure is compatible with the inner-product structure arising from its covariance structure: \begin{equation} \begin{array}{ccc} \text{zero-mean gaussian distribution} & & \text{inner product space} \\ \hline \text{covaraince} & \leftrightarrow & \text{inner ... 1 The distribution of T_{n-1} is independent of \sigma, hence for every value of \sigma the p.value equals the probability of T_{n-1} being larger than |\bar{x} - \mu_0|/(s/\sqrt{n}). Due to symmetry of T_{n-1} around 0, it is suffice to calculate only one sided probaility and multiplying it by 2. 1 HINT 1 : Write the pdf in the form of c*e^{\frac{-1}{2}}(x'\Lambda x) \quad Where \Lambda is symmetric HINT 2: Now invert \Lambda, i.e fine \Sigma = \Lambda^{-1} and then find A such that AA' = \Sigma Then you can argue that if Z follows a standard bivariate normal distribution with mean {0} then X=AZ will also follow bivariate normal ... 1 since there is no inequality sign this is not an accumulating event and therefore its probability is 0? The conclusion is right. The intuition is also basically right, though the expression "accumulating event" is unknown to me. The correct term would be "a zero measure" event. Basically: because U,V are uniform and independent on the unit square, the ... 1 Here is a proof of weak convergence: \sum (a(k)-a_n(k))^{+} \to 0 by DCT because 0 \leq \left(a(k)-a_n(k)\right)^{+} \leq a_k and \sum a_k =1<\infty. But \sum (a(k)-a_n(k))=1-1=0 so we get \sum (a(k)-a_n(k))^{-} \to 0. [Because x^{-}=x^{+}-x]. Adding the two we get \sum |a_n(k)-a(k)| \to 0. It follows easily that \sum_{k\leq j} (a_n(k)-a(k))... 1 There are \binom{10}{5}\times \binom{7}{5}\times 5! different number of ways in which you may select five women, select five men, and choose a way in which to pair them together. Only one of these ways will be correct. To count this, again, choose which five women we suspect are in one of the couples relationships. Then choose which five men we suspect ... 1 Here is an answer that allows n to vary and we look at asymptotics with n. The \Omega(n) property For X a nonnegative random variable with i.i.d. copies \{X_i\}_{i=1}^{\infty}, we say X has the "\Omega(n) property" if E[\max[X_1, ..., X_n]] \geq \Omega(n), that is, if there is a constant c>0 such that:$$ E[\max[X_1,..., X_n]] \geq ... 1 Counterexample: Let it be that$A$has non-degenerate symmetric distribution and that$C=-A$. Then$A\stackrel{d}{=}C$and$A+0=C+2A$(so even stronger than$\stackrel{d}{=}$) but not$0\stackrel{d}{=}2A$1 As mentioned in the comments, it is no more general to state it with$u(X)$where$u$is a non-negative function, or for$X$where$X$is assumed to be non-negative. One way to see this is that the family of all random variables expressible as$u(X)$with the above conditions is seen to coincide with the family of all non-negative random variables. (It is ... 1 There are$\binom{50}{25}$ways to choose$25$people. There are$\binom{30}{15}$ways to choose$15$men, and$\binom{20}{10}$ways to choose$10$men, so$\binom{30}{15}\binom{20}{10}$solutions exist, and a random choice has probability$$\frac{\binom{30}{15}\binom{20}{10}}{\binom{50}{25}}\approx0.227$$of meeting the condition. 1 Since you have only$1$measurement it is meaningless to ask about statistical significance. However, if you have a model of the population distribution of those parameters (for instance, suppose one knows that members of the general population have their glucose level approximately normally distributed with some known mean and standard deviation) then you ... 1 Let me try to answer one interpretation of your question. This should at least be useful to clarify your requirements. Let$Y$be a discrete random variable with range$y_1,\ldots,y_m$and$P(Y=y_j)=p_j$, where$p_j>0.$Let$g$be the "suitable function". For consistency you need$X=g^{-1}(Y)$. Now define for each$y_jn_j$disjoint sets$x_{i,j}\$ ...

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