If X is a binomial random variable, how do you find $E(X(X - 1)(X - 2))$?   Here's my approach:  $E(X(X - 1)(X - 2)) = E(X^3 - 3X^2 + 2X)$  $E(X) = np, Var(X) = np(1 - p)$  $Var(X) = E(X^2) - (E(X))^2$  $E(X^2) = np(1 - p) + (np)^2$  So, $E(X^3 - 3X^2 + 2X) = E(X^3) - 3E(X^2) + 2E(X)$  How should I deal with $E(X^3)$?
120986
Given that $X \sim \operatorname{Binomial}(n,p)$, Find $\mathbb{E}[X(X-1)(X-2)(X-3)]$.   It is suggested that I can transform it into  \begin{align} \mathbb{E}[X(X-1)(X-2)(X-3)]  &amp;=\sum_{k=0}^n k(k-1)(k-2)\mathbb{P}\{X=k\}\\ &amp;=\sum_{k=3}^{n+3} (k-3)(k-4)(k-5)\mathbb{P}\{X=k-3\}\\ &amp;=\sum_{k=0}^n i(i-1)(i-2)\mathbb{P}\{X=i\} \end{align} But then I just have no idea about how can i do it. I suspect that it needs something similar to  but the steps are quite different from this one.  Please help.
6801
If $\mathrm P(X=k)=\binom nkp^k(1-p)^{n-k}$ for a binomial distribution, then from the definition of the expected value $$\mathrm E(X) = \sum^n_{k=0}k\mathrm P(X=k)=\sum^n_{k=0}k\binom nkp^k(1-p)^{n-k}$$ but the expected value of a Binomal distribution is $np$,  so how is      $$\sum^n_{k=0}k\binom nkp^k(1-p)^{n-k}=np$$
97896
Given that the probability density function of random variable $X$ is given by $$f(X)=\begin{cases} \dfrac12,&amp;\text{for} -1\le x\le1\\0,&amp;\text{otherwise}. \end{cases}$$ I tried it using expectations but it turned out that $\Bbb E(X^3)=\Bbb E(X)\Bbb E(X^2)$. What should I do next?
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 $p \rightarrow0 \, \text{ or }\, 1$?
22029
$$P(X_1 &lt; X_2 &lt; X_3)= \frac{(1-p_1)(1-p_2)p_2p_3^2}{(1-p_2p_3)(1-p_1p_2p_3)}$$   To be frank I do not know where to start with this question, I would like an idea to get me going, or better yet an answer. This was on an earlier exam, I would like to be ready if something like this were to come up again tomorrow. What makes this a little bit more difficult than usual is the fact that geometric distribution is  a discrete variable..
51859
Problem: This is problem 33 of section 2.6 (conditioning) in Bertsekas, Tsitsiklis intro to probability text.  We are given that a coin has probability of heads equal to p and tails equal to q and it is tossed successively and independently until a head comes twice in a row or a tail comes twice in a row. What isthe expected value of the number of tosses?  Solution:  Let $X$ be a random value that tracks the number of tosses. In addition, let $H_k, T_k$ be the event that a head or tail comes up on the kth flip respectively. The solution I am looking at involves using total expectation theorem.  I outline the following steps in the solution.  We can partition the sample space into $H_1$ and $T_1$, allowing us to invoke the total expectation theorem. Hence,  $E[X] = pE[X|H_1] + qE[X|T_1]$  Again by total expectation theorem, we can utilize another partition to obtain  $E[X|H_1] = pE[X|H_1,H_2] + qE[X|H_1,T_2] = 2p + q(1 + E[X|T_1])$  $E[X|T_1] = qE[X|T_1,T_2] + pE[X|T_1,H_2] = 2q + p(1+E[X|H_1])$  Stuck: The solution follows up by saying that we can combine the above two relations and use the fact that p+q = 1 to obtain  $E[X|T_1] = \frac{2+p^2}{1-pq} , E[X|H_1] = \frac{2+q^2}{1-pq}$  I have been staring at this for a while now, but I cannot seem to see the steps of the computation. I'd be very grateful if someone could fill in the details regarding the computation. Thank you.
25002
$X_1, X_2, \ldots, X_n$ are $n$ i.i.d. uniform random variables. Let $Y = \min(X_1, X_2,\ldots, X_n)$. Then, what's the expectation of $Y$(i.e., $E(Y)$)?  I have conducted some simulations by Matlab, and the results show that $E(Y)$ may equal to $\frac{1}{n+1}$. Can anyone give a rigorous proof or some hints? Thanks!
72927
Let $\mathbb P(X=1) = \mathbb P(X=-1) = 1/2$. Define       $$X_n = \begin {cases} X &amp; \text{with probability } 1- \frac{1}{n}\\  e^n &amp; \text{with probability }  \frac{1}{n} \end {cases}$$      Does $X_n \leadsto X$ (i.e., converge in distribution)? Does $X_n \stackrel{P}{\rightarrow} X$? Does $\mathbb{E}[(X-X_n)^2] \to 0$?   I think $X_n \leadsto X$ because just by picturing the CDF  as n goes to infinity, their CDFs converge. Can I get some confirmation on this?  But for convergence in probability, I'm having trouble with the limits.   To do this, what I'm trying to do is find $\mathbb P(|X_n-X| \leq \epsilon)$ and then just take 1 minus that as shown on this site:   If $X_n = X$, then $|X_n - X|$ is just $0$ so for all $n$, $|X_n -X| &lt; \epsilon$ But if $X_n = e^n$ then I have the two cases for $X$. For $X=1$, $|X_n-X| = |e^n-1| \leq \epsilon$. and For $X=-1$, $|X_n-X| = |e^n +1| = e^n + 1 \leq \epsilon$. This is where I am stuck. I'm trying to find conditions relating $n$ and $\epsilon$ similiar to here the example on the site. But I'm not sure how to get that from $|e^n - 1| \leq \epsilon$ and  $e^n + 1 \leq \epsilon$ .  Am I even on the right track?
53077
We have the following Random variable:  $Y=\frac{\log Z}{\log Z+\log (1-W)}$ where both $W$ and $Z$ are $U(0,1)$. The question asks to calculate the variance of the random variable $Y$.   I think the random variable $Y$ will follow the exponential distribution with parameter $2$. How I obtained this result is as follows:  First, after a little simplification, the random variable $Y$ turned out to be $\log ZW$ which can further be written as $\log Z+ \log W$. Each of which follows exponential distribution. Hence, the variance of the random variable$Y$ can be obtained as $\frac{1}{4}$. What is wrong in this approach? Because at the back, the answer is $\frac{1}{12}$ as they are saying that $Y$ will follow uniform distribution.
5355
Suppose $X$ is a non-negative random variable satisfying  \begin{align*} P[0 \le X &lt; \infty ]=1. \end{align*} Show   a)  \begin{align*} \lim_{n \to \infty} n E \left( \frac{1}{X}1_{[X&gt;n]} \right) =0 \end{align*} b)  \begin{align*} \lim_{n \to \infty} n^{-1} E \left( \frac{1}{X}1_{[X&gt;n^{-1}]} \right) =0 \end{align*}  What I did  I showed part a) already here is the proof:  \begin{align*} n E \left( \frac{1}{X}1_{[X&gt;n]} \right) \le n E \left( \frac{1}{n}1_{[X&gt;n]} \right)=n\cdot\frac{1}{n}P[X&gt;n]=P[X&gt;n] \end{align*} So $\lim_{n \to \infty} n E \left( \frac{1}{X}1_{[X&gt;n]} \right) \le \lim_{n \to \infty} P[X&gt;n]=0$.  What about part b)? The same techniques no longer works.  I was thinking that  \begin{align*} \lim_{n \to \infty} n E \left( \frac{1}{X}1_{[X&gt;n]} \right) \le \lim_{n \to \infty} n E \left( \frac{1}{X} \right) \le 0 \end{align*} but I am not sure that $E \left( \frac{1}{X} \right) &lt; \infty$
8603
Let $X_1,X_2,X_3,\ldots $ be an i.i.d. sequence of uniform random variables over $[0,1]$. Define $$N= \min \{n \geq 1:X_1+\cdots+X_n&gt;1\}$$ Find $P\{N&gt;n\}$ and compute $E[N]$.
25068
I am working on this problem, where I am required to write down the pgf of $X$, as well as the pgf of $Y$ given $X=j$, and then using conditional expectation find the pgf of $Z$.  So far, I have \begin{align*} G_{X}(z)&amp;=(pz+(1-p))^{n}\\ G_{Y|X=j}(z)&amp;=(qz+(1-q))^{n+j} \end{align*}  I'm new to generating functions and am having difficulty finding the pgf of $Z$. I thought I had a breakthrough when I wrote down \begin{align*} G_{Z}(z)&amp;=\mathbb{E}\left(z^{Z}\right)\\ &amp;=\mathbb{E}\left(z^{n+j-Y}\right)\\ &amp;=\frac{z^{n+j}}{G_{Y}(z)} \end{align*}  But I'm quite sure this isn't correct as I am then required to deduce that $Z$ has the same distribution as the sum of two independent Binomial random variables. I don't know how to apply conditional expectation here. I'm sure this is a very simple problem, and I just want to understand it. Thanks for your patience and time.
55564
Let $X\sim B(n,p_h)$ and $Y\sim B(n,p_\ell)$ be two random variables following a respective binomial distribution, where $p_h&gt;p_\ell$. I want to show that  $$P(X\ge\alpha)\ge P(Y\ge\alpha),$$ for any $\alpha\in\{0,1,\dots,n\}$. In other words, I want to show  $$P(X\ge\alpha)=\sum_{i=\alpha}^{n}\binom{n}{i}p_h^i(1-p_h)^{n-i}\ge P(Y\ge\alpha)=\sum_{i=\alpha}^{n}\binom{n}{i}p_\ell^i(1-p_\ell)^{n-i}.$$  The statement to be proven is obviously true, but proving it is extremely difficult (at least for me). Can anyone give me some guidance on proving it?  It would be nice if someone can give an algebraic proof too, that is, show that  $$\sum_{i=\alpha}^{n}\binom{n}{i}p_h^i(1-p_h)^{n-i}\ge \sum_{i=\alpha}^{n}\binom{n}{i}p_\ell^i(1-p_\ell)^{n-i}.$$  Thank you very much!
88708
Let there be a random variable $X$ and $N(0, \lambda ^2)$ given. I am to find $E[X^n]$ knowing that $\Gamma \big(\frac{1}{2} \big) = \frac{\pi}{2}$.     I started with the definition: $$E[X^n] = \frac{1}{\sqrt{2 \pi}\lambda}\int \limits_{- \infty}^{\infty}x^ne^{\frac{-x^2}{2 \lambda^2}} \mbox{d}x$$ I figured out that: $$E[X^n] = \begin{cases} 0 &amp;\text{for } n = 2k + 1; k \in \mathbb{N} \cup\{0\}\\\text{something}&amp;\text{for } n = 2k; k \in \mathbb{N} \end{cases}$$    So now: $$ \frac{1}{\sqrt{2 \pi}\lambda}\int \limits_{- \infty}^{\infty}x^ne^{\frac{-x^2}{2 \lambda^2}} \mbox{d}x =  \frac{2}{\sqrt{2 \pi}\lambda}\int \limits_{0}^{\infty}x^ne^{\frac{-x^2}{2 \lambda^2}} \mbox{d}x$$ I substitute: $$t = \frac{x}{\lambda}$$ $$\mbox{d}x = \lambda \mbox{d}t$$  Thus: $$\frac{2}{\sqrt{2 \pi}\lambda}\int \limits_{0}^{\infty}x^ne^{\frac{-x^2}{2 \lambda^2}} \mbox{d}x = \frac{2 \lambda^{n}}{\sqrt{2 \pi}}\int \limits_{0}^{\infty}t^ne^{\frac{-t^2}{2}} \mbox{d}t.$$ By integrating the formula above by parts I do get:  $$\frac{2 \lambda^{n}}{\sqrt{2 \pi}}(n-1) \int \limits_{0}^{\infty}t^{n-2}e^{\frac{-t^2}{2}} \mbox{d}t$$ I don't really know where to go from here. And unfortunately I don't see how can I use the Gamma Function, which is given by this formula: $$\Gamma (z) = \int \limits_{0}^{\infty} x^{z-1}e^{-x} \mbox{d}x,$$ isn't it?
83495
Suppose we have three independent exponential random variables $A$, $B$ and $C$  with respective parameters $1$, $2$ and $3$.   Calculate $P(A&lt;B&lt;C)$.  The hint says this problem could be solved with calculus and without calculus. I am really curious how to approach it with different methods.
22586
$X_1$, $X_2$, $X_3$ are independent random variables, each with an exponential distribution, but with means of $2.0, 5.0, 10.0$ respectively. Let $Y$= the smallest or minimum value of these three random variables. Derive and identify the distribution of $Y$. (The distribution function may be useful).    How do I solve this question? Do I plug in each mean to the exponential distribution? I would appreciate it if someone could explain this to me, thanks.
24230
$X$ and $Y$ are independent $U(0,1)$ random variables.   I have to find $E[X|X&gt;Y]$.   I found it out by integrating $\int_0^1 \int_y^1 x dxdy$ to get $\frac{1}{3}$ as the answer but this is wrong according to the answer key. What is the mistake that I have made?