Let $\lambda&gt;0$ be a fixed number. Suppose $X_n\sim Bin(n,\frac{\lambda}{n})$, for $n\in\mathbb{N}\setminus\{0\}$.  Find the moment-generating function $M_{X_n}(t)$ for $X_n$. Compute for $t$ fixed, $\lim_{n\rightarrow\infty}M_{X_n}(t)$. What is the distribution of the m.g.f. that corresponds to the limit?  My Attempt  $\displaystyle{M(t)=\mathbb{E}[e^{tx}]=\sum_{x=0}^xe^{xt}\binom{n}{x}\left(\frac{\lambda}{n}\right)^x\left(1-\frac{\lambda}{n}\right)^{n-x}=\sum_{x=0}^x\binom{n}{x}\left(\frac{\lambda e^{t}}{n}\right)^x\left(1-\frac{\lambda}{n}\right)^{n-x}=\left(\frac{\lambda e^t}{n}+1-\frac{\lambda}{n}\right)^n}$  \begin{align} \lim_{n\rightarrow\infty}\left(\frac{\lambda e^t}{n}+1-\frac{\lambda}{n}\right)^n&amp;=\lim_{n\rightarrow\infty}\left(1+\frac{\lambda e^t-\lambda}{n}\right)^n\\ &amp;=\lim_{n\rightarrow\infty}e^{n\ln(1+\frac{\lambda e^t-\lambda}{n})}\\ &amp;=\lim_{n\rightarrow\infty}(\lambda e^t-\lambda)+\frac{o(n)}{n}\quad\text{(where $\lim_{n\rightarrow\infty}\frac{o(n)}{n}=0$)}\\ &amp;=\lambda e^t-\lambda \end{align}   The part that I am having trouble with is figuring out which distribution the limit of the m.g.f. corresponds to.
123606
$X_n = \operatorname{Bin}(n,\frac{\lambda}{n})$ for $\lambda &gt; 0$. Show $M_{X_n}(t)$ converges to the MGF of Poisson($\lambda)$ as $n\rightarrow \infty$  So far I got MGF of $$\operatorname{Poi}(\lambda)=e^{-\lambda}\sum_{x=0}^\infty(e^tx)^x/x!$$ MGF of $$\operatorname{Bin}(n,\frac{\lambda}{n})=\sum_{x=0}^{\infty}e^{tx}{n\choose x}(\frac{\lambda}{n})^x(1-\frac{\lambda}{n})^{n-x}$$ $$\lim_{n\to\infty} \sum_{x=0}^{\infty}e^{tx}{n\choose x}(\frac{\lambda}{n})^x(1-\frac{\lambda}{n})^{n-x}$$  I can see that $n \choose x$ as n goes to infinity will be $\frac1{x!}$, but I don't know how they got the $e^{-\lambda}$ and $x^x$. I could be wrong but I feel like $(1-\frac{\lambda}{n})^{n-x}$ just becomes 1 and $(\frac{\lambda}{n})^x$ becomes 0, which wouldn't really work out.
12903
Suppose X ∼ Exponential(λ). Find the PDF of $$Y = sin^2(X)$$ Note: While the solution involves an infinite sum, it is an example of a geometric series and so can be evaluated.  My answer so far: I was able to go as far as finding the bounds, and now I am stuck on the integral itself. As in the directions, we are told to use a geometric series, but I don't know how to proceed from here.   The integral which will give me the CDF   I also have the piece $$P(0 &lt;= x &lt;= arcsin(\sqrt t))$$ I know this does not included in the integral, but I don't know how to incorporate it into my CDF.   $$\int_a^b \lambda\ e^-\lambda x\ \,dx,$$  $$ a = k \pi\ - arcsin(\sqrt t)$$  $$ b = k \pi\ + arcsin(\sqrt t)$$
118521
Let $X \sim Bin(n, p)$ and let   $$\hat p = \frac{X}{n} $$ be an estimator of the probability $(p)$ of success in this binomial distribution, where $X$ is the number of successes and $n$ is the number of trials.  My question is, why does $\text{MSE}(\hat p)\rightarrow 0$ as $p \rightarrow0 \, \text{ or }\, 1$?
32897
I have two questions.   1) If $X_1,X_2,X_3,...,X_n$ constitute a random sample of size $n$ from an exponential distribution, show that $\bar X$ is a consistent estimator of the parameter $\lambda$.    This is my attempt: Mean of exponential distribution is $\lambda^{-1}$. Thus by central Limit theorem $E(\bar X)=\lambda^{-1}$. Using mean square consistency, if  $\lim_{n\to \infty} E[(\bar X -\lambda )^2]=0$ then $\bar X $ is consistent for $\lambda$. Since $$ E[(\bar X -\lambda )^2]= E[(\bar X)^2]+E(\lambda)^2-2\lambda E(\bar X) , $$  I get $$ \lim_{n\to \infty} E[(\bar X -\lambda )^2] =    \lim_{n\to \infty} {2 \over \lambda ^2 }+\lambda^2-2 . $$ Clearly this is not equal to $0$. How can I show this is consistent? I tried using Chebechev inequality as well. But even using that I couldn't.    2) Is $X_{(n)}$ a consistent estimator for $\lambda$? If $P[|X_{(n)}-\lambda|&lt;\epsilon]$ approaches $1$ as $n$ goes to infinity for all $\epsilon &gt; 0$ then $X_n$ is a consistent estimator for $\lambda$.    The probability density function of $X_{(n)}$ is $f_n(x) = n \lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}$,  for $X&gt;0$.  Thus \begin{align*} P[|X_{(n)}-\lambda|&lt;\epsilon]  &amp;= P[\lambda -\epsilon&lt;X_{n}&lt;\lambda+\epsilon] \\  &amp;=P[0&lt;X_{n}&lt;\lambda+\epsilon] \\  &amp;=\int_0^{\lambda+\epsilon} n \lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}\\  &amp;= (1-e^{- \lambda X_{(n)} }), \end{align*} where $X{(n)}$ takes values in $[0 , \lambda+ \epsilon]$.  Thus, can I say $\lim_{n\to \infty}[1-e^{-\lambda(\lambda+\epsilon)}]^n=1$?
33243
Let $P(k,X)=p(1-p)^{k-1}$.  When deriving the moment generating function I start off as follows:  $E[e^{kt}X]=\sum\limits_{k=1}^{\infty}e^{kt}p(1-p)^{k-1}$.  How I end up rearranging this is as follows:  $\frac{p}{1-p}\sum\limits_{k=1}^{\infty}e^{kt}(1-p)^k=\frac{p}{1-p}\sum\limits_{k=1}^{\infty}(e^{t}(1-p))^k=\frac{p}{1-p}\frac{1}{1-e^t(1-p)}$  I'm obviously not arriving at the correct answer, but I want to know why this derivation is wrong.
70965
I would very much appreciate a hint for the following problem  Let $\left(X_n\right)_{n=1}^\infty$ be a sequence of random variables s.t.: $$X_1 \sim U_{[0,1]}$$ and for all $n&gt;1$: $$X_n \sim U_{[0,X_{n-1}]}.$$ Give a general expression for $f_{X_n}$ the pdf of $X_n$.  Thanks
93756
We are given $R(t)$ = $P(X&gt;t)$ for all $x &gt; 0$ and  $$R(0) = 1 - Fx(0) = 1\text{ and }\lim\limits_{t \to \infty} R(t) = 0$$  The random variable $X$ also satisfies the memoryless property:  $$P(X&gt;s+t|X&gt;t) = P(X&gt;s)\text{ for }s&gt;0\text{ and }t&gt;0$$  Let $R&#39;(0) = - \lambda\ where \  \lambda\ ,$  is some positive constant. I need to show that X must be exponentially distributed.  Given that $\dfrac{R(t + h) - R(t)}{h}$ = $R(t)\left[\dfrac{R(h) - 1}{h}\right]$   Show that by letting $\lim\limits_{h\to \infty}$ $\dfrac{dR(t)}{dt} = -\lambda$$R(t)$  (I think we should use Hopital's rule here I am not sure by differentiating $\left[\dfrac{R(h) - 1}{h}\right]$ and letting $h$ tend to $0$, we will get $-\lambda$ for this part but I got stuck afterwards).  Also argue that $X$ is an exponential random variable with rate parameter $\lambda$  by solving the differential equation above respecting the conditions:  $$R(0) = 1 - Fx(0) = 1\text{ and }\lim\limits_{t \to \infty} R(t) = 0$$
37386
I have a three-dimensional random vector $Y$ ~ $N(\mu , \Sigma)$ whose moment-generating function is:  $M_Y(t_1,t_2,t_3)=exp(2t_1+3t_2-3t_3+4t_1^2+5t_2^2+3t_3^2-t_1t_2+2t_1t_3)$.  It was easy to calculate $\mu$  because $M_Y(t)=exp(\mu^Tt+\frac12 t^T\Sigma t), \mu_1 = 2, \mu_2 = 3, \mu_1 = -3.$  Also, $\Sigma=\left[\matrix{8&amp;-1&amp;2 \\ -1&amp;10&amp;0 \\ 2&amp;0&amp;6}\right]$. (If there was any wrong points in this calculation, please let me know).  This is what I have done now....  However, I know only how to calculate the conditional mean when the vector is two-dimentional: $E[Y_1|Y_2=y_2]=E[Y_1]+\Sigma_{12}\Sigma_{22}^{-1}(y_2-\mu_2)$  $VAR[Y_1|Y_2=y_2]=\Sigma_{11}-\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}$ (* May anyone let me know the process of having the inverse of covariance in the $\Sigma$?)  For 3D, I still did not get good idea how to calculate $E[Y_1|Y_2=y_2, Y_3=y_3] $ and $VAR[Y_1|Y_2=y_2, Y_3=y_3]$.  For example: $E[Y_1|Y_2=\frac72, Y_3=-2] $ and $VAR[Y_1|Y_2=\frac72, Y_3=-2]$. How is the process changing compared to a 2D one, and where should I start?
105717
Let $X_1$ and $X_2$ be independent Normal random variables with mean $\mu_1$ and $\mu_2$, and variances $\sigma_1$ and $\sigma_2$. Let $Y = X_2-X_1 + c$, where $c$ is a constant. For notational simplicity, let $Y = X' + c$, where $X' = X_2-X_1$ The moment generating function of a Normal random variable $X_n\sim N(μ_n,σ_n^2)$ is $$M_{X_n } (t)=\text{exp}\Big(μ_n t+\frac{σ_n^2 t^2}{2}\Big)$$ Then, $$M_{X'} (t)=\text{exp}\Big[t(\mu_1+\mu_2)+\frac{t^2}{2}(\sigma^2_1+\sigma^2_2)\Big]$$ Thus, by the uniqueness property of moment generating function, $X'$ is said to follow a Normal distribution with mean $(\mu_1+\mu_2)$ and variance $(\sigma^2_1+\sigma^2_2)$. How do I proceed from here to derive the following:  First, I want to establish that $Y$ is also a Normal random variable using the moment generating function. The constant $c$ bugs me, but I cannot ignore it. As per my understanding, $Y\sim N\big(c+\mu_1+\mu_2,\sigma^2_1+\sigma^2_2\big)$ (and I could be completely wrong). But I don't understand how to apply the moment generating function to prove that.  Secondly, I want to derive $P(Y \leq a)$, where $a$ is a constant. If I am not mistaken, $P(Y \leq a)$ can be written as $$P(Y \leq a)=\Phi\Bigg(\frac{a-X_2-X_1 + c}{\sqrt{2(\sigma^2_1+\sigma^2_2)}}\Bigg).$$ But how do I arrive at that.   Any help is high solicited and appreciated.
117265
This is an exercise of the Central Limit Theorem:     Let $Y^{\lambda}$ be a Poisson random variable with parameter $\lambda&gt;0$.   Prove that $\frac{Y^{\lambda}-\lambda}{\sqrt{\lambda}}\to Z\sim N(0,1)$ in distribution as $\lambda\to\infty$.   I've done that  $$ Z_n\to Z\sim N(0,1) $$ in distribution using the CLT, where $Z_n=(Y^n-n)/\sqrt{n}$. Some naive attempt to go is considering  $$ Y^{n}\leq Y^{\lambda}\leq Y^{n+1}\tag{*} $$ where $n\leq\lambda\leq n+1$ and somehow use the squeeze theorem. But both (*) and the squeeze theorem in convergence in distribution are NOT justified. How can I go on? Or do I need an alternative direction?
103006
I  know that to genarete ramdom variables following exponential distribution just do:  $$X=-\frac{1}{\lambda}ln(U)$$  where $U\sim U(0,1)$  Now, to find a distribution restricted to the interval $(a, b)$, I used the truncated distribution () coming in the following cumulative distribution  $$H(t)=\frac{e^{-\lambda a}-e^{-\lambda t}}{e^{-\lambda b}-e^{-\lambda a}}$$  However I don't Know to find the inverse function $H^{-1}$. Also I don't know if this expression is correct. I thank
30668
The classic example... $X \sim Po\left (\lambda\right ), Y \sim Po\left (\mu\right)$, X and Y are independent.  Show that the conditional distribution of X is binomially distributed. Or in other words, $P(X=k\mid X+Y = n) = P (\tilde{X} = k), \tilde{X} \sim B\left (n ,\frac{\lambda}{\lambda + \mu}\right )$.   I've so far managed to reach to this step, and have been stuck since. Just somehow gotta get a $\frac{1}{n!}$ in the denominator, that would then complete the proof..or at least I think..  $$P(X=k\mid X+Y=n) = \frac{\frac{\lambda^{k}\mu^{n-k}}{k!(n-k)!}}{P(X+Y = n)}=  \frac{\frac{\lambda^{k}\mu^{n-k}}{k!(n-k)!}}{\sum_{i=1}^{n} \frac{\lambda^{i}\mu^{n-i}}{i!(n-i)!}}$$  Thanks for the help!
53341
Let $X_1, X_2, \dots$ be i.i.d. exponentially distributed RVs. For $n = 1,2,\dots$ consider: $Y_n := \max(X_1, X_2, \dots, X_n)$ $U_n := \sum_{i=1}^{n}\frac{X_i}{i}$ Show that $Y_n$ and $U_n$ have the same distribution   What I've tried: $P(Y_n&lt;y)= P(X_i&lt;y)^n = (1- e^{\lambda y})^n =&gt; P(Y_n=y) = n(1- e^{\lambda y})^{n-1}$ But I get stuck with $U_n$. I tried an MGF, $U_n$ evaluates nicely but then $Y_n$ gets messy. Any thoughts?
83989
The probability $P(X=n)$ that an event X takes place $n$ times in a fixed period of time follows the Poisson distribution with parameter $\lambda$ i.e.  $$ P(X = n) = e^{-\lambda} \frac{\lambda ^ n}{n!}$$  I have to evaluate the probability that the event $X$ takes place an even number of times. I know that:   $$ P(X \text{ is even} ) = e^{-\lambda} \cdot \sum_{k=0}^{+\infty} \frac{\lambda^{2k}}{(2k)!}$$  but I can't solve the series.  I guess that I have to use the fact that $e^\lambda = \sum_{n = 0}^{+\infty} \lambda^n/n!$, but I got stuck.  How can I evaluate $P(X \text{ is even})$ (alternative solutions appreciated).
122532
Let $Y_1, \ldots, Y_n$ be a random sample from the pdf $$f(y\mid\theta) = \theta y^{\theta - 1}, 0 &lt; y &lt; 1, \theta &gt; 0 .$$ show that $$E\left(\frac{1}{2\theta\sum_{i = 1}^n W_i}\right) = \frac{1}{2(n-1)}$$  I have showed that $\sum_{i = 1}^n -\ln(Y_i)$ is sufficient for $\theta$, and if $W_i = -\ln(Y_i)$, I have showed that $W_i$ has an exponential distribution with mean $1/\theta$. I have also showed that $2\theta \sum_{i=1}^n W_i $ has a $\chi^2$ distribution with $2n$ degrees of freedom.  I am wondering how to show it?
97524
Let $T_n$ be a random variable with $T_n=X_1+...+X_n$ where the $X_i$'s are iid. Further we set $N(t)=max\{ n: T_n\leq n\}$ with the property $\Pr(N(t)&lt;\infty)=1$.  I want to prove that $\mathbb{E}[N(t)]&lt;\infty$.   I have done this so far. Let $F_n$ be the distribution function of $T_n$. $$\mathbb{E}[N(t)]=\sum_{n=1}^{\infty} n P(N(t)=n)=\sum_{n=1}^{\infty} n (F_n(t)-F_{n+1}(t))=\sum_{n=1}^{\infty} F_n(t)$$  Further I know that $0 \leq F_n \leq 1$. But does this implies $\mathbb{E}[N(t)]&lt;\infty$ ? I don't think so... I don't know what I can do further...   Can anybody help me how I can proof it?   Thank you for any help!!!
3657
Let $X$~$\mathsf{Poisson}$$(rate=\lambda_1)$ and $Y$~$\mathsf{Poisson}$$(rate=\lambda_2)$. Let $Z=X+Y$. Find the distribution of $X|Z=n$ and name it.  Here's what I tried:  $$\begin{align}\mathsf{M_{X|X+Y=n}}(t) &amp;\stackrel{?}{=} \mathsf{E}[e^{tX}|X+Y=n]\tag 1 \\[1ex]&amp;= \mathsf{E}[e^{tX}|X=n-Y]\tag 2 \\[1ex]&amp;\stackrel{?}{=} \mathsf{E}[e^{t(n-Y)}] \tag 3\\[1ex]&amp;= \mathsf{E}[e^{tn}e^{-tY}] \tag 4\\[1ex]&amp;\stackrel{?}{=} e^{tn}\mathsf{E}[e^{(-t)Y}]\tag 5 \\[1ex]&amp;= e^{tn}\mathsf{M_Y}(-t)\tag 6 \\[1ex]&amp;= e^{tn} \cdot e^{\lambda_2(e^{-t}-1)}\tag 7\end{align}$$  And this is where I am stuck. I don't see how I can work the $tn$ in with the other exponent in such a way that I have a moment generating function that I recognize as the MGF of a distribution I know. I am also uncertain about some of the steps I took to get this MGF, so I put question marks over those equals signs. Lastly, my homework does not include the information that $X$ and $Y$ are independent, but if this information is necessary to make the problem workable then I intend to assume that my instructor forgot to include it. Thanks for the help.