As the title states, how does $E(|X|)=\int_0^{\infty}P[|X|\ge x]dx$ ? The only assumption being that $E(|X|)\le \infty$  Mybe I can use the identity function in some way, since $E[1_{X\ge x}]=P[X\ge x]$?  Thanks in advance!
24512
Let $X$ be a non-negative random variable and $F_{X}$ the corresponding CDF. Show,   $$E(X) = \int_0^\infty (1-F_X (t)) \, dt$$   when $X$ has : a) a discrete distribution, b) a continuous distribution.   I assumed that for the case of a continuous distribution, since $F_X (t) = \mathbb{P}(X\leq t)$, then $1-F_X (t) = 1- \mathbb{P}(X\leq t) = \mathbb{P}(X&gt;  t)$. Although how useful integrating that is, I really have no idea.
63223
Given the integral $\int_{0}^{\infty} \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})} dx$  How can I evaluate it? Thanks!
98938
Let $X$ be a non-negative integer-valued random variable with finite mean. Show that $$E(X)=\sum^\infty_{n=0}P(X&gt;n)$$  This is the hint from my lecturer.  "Start with the definition $E(X)=\sum^\infty_{x=1}xP(X=x)$. Rewrite the series as double sum."  For my opinion. I think the double sum have the form of $\sum\sum f(x)$, but how to get this form? And how to continue?
116635
Let's say - $F(x) &gt; G(x)$ for all $x$ in $E$.  So is it true that ?  $$\int F(x)\,dx  &gt; \int G(x)\,dx$$ or I can use this statement only when the two functions are non-negative functions.  Thanks!
40488
Show that for a non-negative random variable $X$,   $$\mathbb E(X)=\int_0^\infty \mathbb P(X\ge x) \, dx.$$   I started with  $$\mathbb E(X)=\int_0^\infty x \, dF(x)=\int_0^\infty \int_0^x dt\,dF(x).$$ This is my try.
27738
Prove that      $$ \int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}\,dx = \ln\left(\frac{a}{b}\right) $$   My Attempt:  Define the function $I(a,b)$ as  $$ I(a,b) = \int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}\,dx $$  Differentiate both side with respect to $a$ to get  $$ \begin{align} \frac{dI(a,b)}{da} &amp;= \int_{0}^{\infty}\frac{0-e^{-ax}(-x)}{x}\,dx\\ &amp;= \int_{0}^{\infty}e^{-ax}\,dx\\ &amp;= -\frac{1}{a}(0-1)\\ &amp;= \frac{1}{a} \end{align} $$  How can I complete the proof from here?
75972
I am wondering if the following expectation can be computed in the shown manner  $$\begin{aligned} E[\log_2(1+X) \mid X&gt;a] &amp;= \int_{0}^{\infty} P( \log_2(1+X)&gt;t \mid X&gt;a) dt \\ &amp;= \int_{0}^{\infty} P(X&gt;2^t-1 \mid X&gt;a) dt \\  &amp;= \int_{0}^{\infty} \frac{P(X&gt;2^t-1 \cap X&gt;a)}{P(X&gt;a)} dt \\ &amp;= \frac{1}{P(X&gt;a)} \left(\int_{0}^{a} P(X&gt;a) dt + \int_a^{\infty} P(X&gt;2^{t}-1) dt \right)  \end{aligned}$$  Note that there is no closed form expression for the PDF of X hence I am trying to find alternative ways to compute the expectation.  Thank you
56322
Given a random variable $X\sim\mathcal N(0,\sigma^2)$, how can we prove that $E[X^4]=3\sigma^4$? I am having trouble even starting with the proof.
101743
Studying some probability theory and I came across this question. Show that if $Y$ is a non-negative random variable, and $p&gt;0$, $$E(Y^p)=\int_{0}^{\infty}px^{p-1}P(Y\geq x)dx$$ I'm a bit stuck on this question: I know that by Markov's inequality, $\frac{E(Y^p)}{y^p}\geq P(Y&gt;y)$ since $Y=\lvert Y\rvert$ here. Also, $y^p=\int_{0}^y px^{p-1}dx$ and I believe these two facts can be used to solve the problem, but I'm not sure how exactly.  Could anyone give me some help, point me in the right direction?
58203
I stumbled on this question while answering the following question:  Given a general wavefunction $\psi(x)$ and an operator $\hat Q$, with eigenvalue spectrum $\{q_r\}$ and corresponding eigenfunctions $\{u_r\left(x\right)\}$, show that the expectation value of $\hat Q$ is given by $$\langle\hat Q\rangle = \int_{-\infty}^\infty{\psi^*\hat Q \psi \,dx}$$  I started of with the ususal definition of the expected value and inserted the identity operator in position representation: $$\langle\hat Q\rangle = \langle \psi|\hat Q|\psi\rangle = \langle \psi|\left(\int dx|x\rangle\langle x|\right) \hat Q\left(\int  {dx}'|x'\rangle\langle x'|\right)|\psi\rangle$$  Taking $\langle\psi|$ and $|\psi\rangle$ into the integral gives: $$\left(\int dx\langle \psi|x\rangle\langle x|\right) \hat Q\left(\int  {dx}'|x'\rangle\langle x'|\psi\rangle\right)$$  Pulling the 2nd integral into the first:$$\int dx\int  {dx}'\langle \psi|x\rangle\langle x'|\psi\rangle\langle x| \hat Q|x'\rangle$$  This is the point where I'm stuck. If $\langle x|\hat Q|x'\rangle = \hat Q \delta(x-x')$, then the last expression would evaluate to the answer. However I  don't know why this should be true and even if it is true I wouldn't know how to prove this equality.
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?
45342
I wish to use the  to calculate the variance of a normal-distributed function. For this, I need the expected value of $X$ as well as the one of $X^2$. Intuitively, I would have assumed that $E(X^2)$ is always equal to $E(X)^2$. In fact, I cannot imagine how they could be different.  Could you explain how this is possible, e.g. with an example?
103005
I am looking for conditional expectation $E[X| X&gt;Y]$ for independent $X, Y \sim N(0,1).$ I just showed that $P(X\leq a | X&gt;Y) = \Phi(a)^2$, where $\Phi(a)$ is CDF for Normal$(0,1)$, but I don't know how to compute integral in order to obtain expected value.
118361
Let T be a non-negative integer-valued random variable with $\mathbb{E}(T) &lt; \infty $. Prove that $\mathbb{E}(T) = \sum^\infty_{k=1}\mathbb{P}(T \geq k)$.  Had a few attempts, haven't really got anywhere. I'm wondering as I'm typing this if proof by induction is a good way to go.  Edit: One major thing I forgot to add, am I correct in thinking that also, $\mathbb{E}(T) = \sum^\infty_{k=1}k\mathbb{P}(T = k)$?
71185
I am currently trying to understand how does an instance of the EM algorithm is derived when there are "hidden variables" in this old paper: .  I don't really understand how the expectation step is derived, and what bothers me the most, is that I cannot prove the following statement (right before 2.11):  $$ \mathbb{E}[X|X+Y] = \frac{(X+Y)\lambda_X}{\lambda_X+\lambda_Y} $$  where $X \sim Poisson\{\lambda_X\}$ and $Y \sim Poisson\{\lambda_Y\}$  To begin, I wrote the discrete expectation as follows: $$ \mathbb{E}[X|X+Y] = \sum_{k=1}^{X+Y} k P(X=k|X+Y)\\ = \sum_{k=0}^{X+Y} k \frac{P(X+Y|X=k)P(x)}{P(X+Y)} $$  Where I can replace $$ P(X+Y=k) = \frac{(\lambda_X+\lambda_Y)^ke^{-(\lambda_X+\lambda_Y)}}{k!}$$ $$ P(k=x) = \frac{\lambda_X^xe^{-\lambda_X}}{x!}$$ $$ P(X+Y=k|x) = P(Y=x-k) = \frac{\lambda_Y^{x-k}e^{-\lambda_Y}}{(x-k)!}$$  but I dont really know how to mix that together, because I don't really see what does the event $X+Y$ stands for, and how does it rely on the x in the integral ?  Thank you very much for you help
96591
Given that a random variable $X$ is a normally distributed variable with mean $\mu$ and variance $\sigma^2$. Why exactly is:  $$\mathbb{E}[|X|] = \mathbb{E}[X\mid X&gt;0]P(X&gt;0) - \mathbb{E}[X\mid X&lt;0]P(X&lt;0)$$  Because we have that:  $$\mathbb{E}[X\mid X&gt;0] = \frac{\int_0^\infty xf_X(x)\,dx}{P(X&gt;0)}$$  Therefore my definition immediately above must be incorrect?  Because if it is not we have that:  $$\mathbb{E}[|X|] = \int_0^\infty xf_X(x)\,dx - \int_{-\infty}^0 xf_X(x)\,dx$$  and where does the minus there come from anyway, and why?
78884
I am trying to simplify $E[YE[Y|X]|X]$ can I use this property:   $$E[E[Y|X]|X]= E[Y|X]$$  If yes I have never seen a Proof of this property (that seems very reasonable), could I have a reference? If I am right the correct result would be $$E[YE[Y|X]|X] = E[(Y|X)]^2$$