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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 example $$P( 4 < X < 11 ) = P(A) = P( \{ 4 < X < 11\}),$$ 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).


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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 $\mathbb{E}[X]=1$.


1

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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