Let $X$ be a real-valued random variable with the cdf $F_X$. I need to show that $$ E(X)=\int_0^\infty (1-F_X(x))dx-\int_{-\infty}^0 F_X(x)dx $$
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.
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?
55308
I am wondering if there is such a sequence of random variables $(X_n)_{n=0}^\infty$ such that $\mathbb{E}(X_{n+1}\mid X_n)=X_n$ for all $n\geq0$ but which is not a martingale with respect to the filtration $\mathcal{F_n}=\sigma(X_0, \dots, X_n)$.  I would really appreciate if you could show me such an example.
49961
Let $X_1$ and $X_2$ be two independent exponential random variables with the PDFs $f_{X_1}(x_1)=\lambda_1 \exp(-\lambda_1 x_1)$ and $f_{X_2}(x_2)=\lambda_2 \exp(-\lambda_2 x_2)$.   Let $X= \frac{X_1X_2}{X_1+X_2+a}$ be a r.v. for which I want to derive the CDF, where $a$ is a positive constant.  In other words, I need to calculate $\mathbb{P}\{X &lt;x \}$.  Any ideas or hints?  Edit (attempt):  Using a similar approach to that used , we have $X_1 \in [0, \infty)$ and  $X_2 \in [0, \frac{x(x_1+a)}{x_1-x})$. The latter interval results from: $\frac{x_1x_2}{x_1+x_2+a} &lt; x$, thus $x_2(x_1-x)&lt;x(x_1+a)$ which implies $x_2 &lt; \frac{x(x_1+a)}{x_1-x}$; but I am not sure if this is correct since $x_1-x$ can sometimes be negative (?). Based on the above, we can write: \begin{align}  \mathbb{P} \{ \frac{X_1X_2}{X_1+X_2+a} &lt; x \}  &amp; = \int_{x_1=0}^\infty \int_{x_2=0}^{ \frac{x(x_1+a)}{x_1-x}} \lambda_1 \exp(-\lambda_1 x_1) \lambda_2 \exp(-\lambda_2 x_2)  dx_2 dx_1  \\ &amp;= \int_{x_1=0}^\infty \left(1-\exp(-\lambda_2\frac{x(x_1+a)}{x_1-x}) \right) \lambda_1 \exp(-\lambda_1 x_1) dx_1  \\ &amp; = 1-\lambda_1 \int_{x_1=x}^\infty \exp(-\lambda_2\frac{x(x_1+a)}{x_1-x} ) \exp(-\lambda_1 x_1)   dx_1 \end{align} In the last equality: for the integral of $\lambda_1 \exp(-\lambda_1 x_1) dx_1$, I suppose that $x_1 \in [0,\infty)$, whereas for the second term I suppose that $x_1 \ge x$ (since $x_2$ should be positive) (?).  Is my approach correct ?
92968
In general can one say that for a random variable X:  $E[\frac{1}{X}] = \frac{1}{E[X]}$ ?  I've worked out a few examples where this works but I'm not sure how widely this is useful...
40132
Let $f : \mathbb{R} \to [ 0, \infty)$ be a measurable function. If $\int_{- \infty}^{\infty} f(x) dx =1$. Then I want show that $ \int_{- \infty}^{\infty}  \frac{1}{1+ f(x)} dx = \infty.$  Any help will be appreciated.  Thank you in advance.
116254
if $x_i$ are IID  continuous random variables, with $E[x_i]=μ$, is the following correct?   $E\left[\prod_{i=1}^n(1+⍺\space x_i)^i\right] = \prod_{i=1}^nE\left[(1+⍺\space x_i)^i\right] =  \prod_{i=1}^n(1+⍺\space E\left[x_i\right])^i = (1+⍺\space μ)^n$  thanks!
4583
$\def\d{\mathrm{d}}$Given that $X_1, X_2, \cdots$ are i.i.d., prove that$$ \frac{1}{n} \max_{1 \leqslant k \leqslant n} |X_k| \xrightarrow{\mathrm{a.s.}} 0 \Longleftrightarrow E(|X_1|) &lt; +\infty. $$   My try:  Because $\varliminf\limits_{n \to \infty} \dfrac{1}{n} \max\limits_{1 \leqslant k \leqslant n} |X_k|$ and $\varlimsup\limits_{n \to \infty} \dfrac{1}{n} \max\limits_{1 \leqslant k \leqslant n} |X_k|$ are both measurable with respect to the exchangeable σ-algebra of $X_1, X_2, \cdots$, they are both almost surely constants by the Hewitt-Savage 0-1 law, i.e.$$ \varliminf_{n \to \infty} \frac{1}{n} \max_{1 \leqslant k \leqslant n} |X_k| = c_1, \quad \text{a.s.} \tag{1}\\ \varlimsup_{n \to \infty} \frac{1}{n} \max_{1 \leqslant k \leqslant n} |X_k| = c_2, \quad \text{a.s.} $$ where $c_1, c_2 \in [0, +\infty]$. Also, if the distribution function of $|X_1|$ is $F(x)$, then the distribution function of $\dfrac{1}{n} \max\limits_{1 \leqslant k \leqslant n} |X_k|$ is$$ F_n(x) = P\left( \frac{1}{n} \max_{1 \leqslant k \leqslant n} |X_k| \leqslant x \right) = \prod_{k = 1}^n P(|x_k| \leqslant nx) = (F(nx))^n, $$ which implies\begin{align*} E\left( \frac{1}{n} \max_{1 \leqslant k \leqslant n} |X_k| \right) &amp;= \int_0^{+\infty} x \,\d((F(nx))^n) = \frac{1}{n} \int_0^{+\infty} y \,\d((F(y))^n)\\ &amp;= \frac{1}{n} \int_0^{+\infty} y·n (F(y))^{n - 1} \,\d F(y) = \int_0^{+\infty} y (F(y))^{n - 1} \,\d F(y)\\ &amp;\leqslant \int_0^{+\infty} y \,\d F(y) = E(|X_1|). \end{align*} Also for any $c &gt; 0$,$$ \sum_{k = 1}^\infty P(|X_1| \geqslant ck) &lt; \infty \Longleftrightarrow E(|X_1|) &lt; +\infty. \tag{2} $$ However, I failed to connect (1) and (2) to prove the proposition.  Any ideas are appreciated. Thanks in advance.
94593
Let $X_1, X_2, \dots, X_n$ be $n$ positive iid random variables. Then show that $$E\left(\frac{\sum_{j=1}^k X_j}{\sum_{i=1}^{n} X_i}\right) = \frac{k}{n}.$$  Because of the linearlity of the expectation I known that $E\left(\frac{\sum_{j=1}^k X_j}{\sum_{i=1}^{n} X_i}\right) = \sum_{j=1}^k E\left(\frac{X_j}{\sum_{i=1}^{n} X_i}\right)$, so it's enougth to show $E\left(\frac{X_1}{\sum_{i=1}^{n} X_i}\right) = \frac{1}{n}$. But I'm unable to deal with the $X_i$ in the denominator.
11201
Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, and let $X$ be a nonnegative random defined on this space. For any $A\in \mathcal{F}$, let $\mathbb{E}[X;A]:=\mathbb{E}[X\mathbb{1}_{A}]$, where $\mathbb{1}_{A}$ denotes the indicator random variable on $A$. Without assuming $\mathbb{E}\left[\frac{1}{X}\right]&lt;\infty$, show that  \begin{eqnarray} \lim\limits_{y\downarrow 0} ~~y~\mathbb{E}\left[\frac{1}{X};X&gt;y\right]=0. \end{eqnarray} Any initial ideas will be greatly appreciated.
24993
I don't really know how to start proving this question.  Let $\xi$ and $\eta$ be independent, identically distributed random variables with $E(|\xi|)$ finite.  Show that $E(\xi\mid\xi+\eta)=E(\eta\mid\xi+\eta)=\frac{\xi+\eta}{2}$  Does anyone here have any idea for starting this question?
682
Problem: Let X be a nonnegative random variable and $log X \sim N(0,1)$, prove that $\mathbb{E}[X^k] = e^{\frac{k^2}{2}}$.  I'm self learning probability in order to understand some theorems in number theory, and this problem came in a book I'm reading.   I know that $\mathbb{E}[X^k]=\mathbb{E}[e^{klog X }]=\int_{-\infty}^{\infty} e^{klog(x)}f_{X}(x)$.  Now, I'm pretty lost in the $f_{X}(x)$ part. I suppose the question is how can I link the logX which its distribution is known and the X but Im not sure what is allowed to use in these cases.   Any help would be really appreciated.   Thanks! :)
50765
I have a random variable $X(a) = \log(a)$ where a is normal distributed $\mathcal N(\mu,\sigma^2)$. What can I say about $E(X)$ and $Var(X)$? An approximation would be helpful too.
89896
The random variable $X$ is exponentially distributed with parameter $\lambda = \dfrac{1}{2}$. Also, define $Y = -2X + 1$. Calculate $E(Y)$ and $Var(Y)$.  So we have   $$f_X(x) = \dfrac{1}{2} e^{-\dfrac{1}{2}x} $$  if $x\ge0$ and $0$ otherwise.  But how to calculate $E(Y)$ and $Var(Y)$ ?
57272
This is my first post so I will try to be as clear and concise as possible. I am doing a course in statistics and we define the true mean of a random gaussian variable to be as follows: $\mu$ = $\int_{R}xdF_{N(\mu, \sigma^2)}(x)$ As the Lesbesgue-Stieljes integral. In the notes we defined the empirical CDF as follows: ${F_{n}}$ = $\frac{1}{n}\sum_{i=1}^{n}\mathbb{1}_{\{X_{i} \le x \}}$ We then proceed to show that: $\mu$ = $\int_{R}xd(\frac{1}{n}\sum_{i=1}^{n}\mathbb{1}_{\{X_{i} \le x \}})$ = $\frac{1}{n}\sum_{i=1}^{n}X_{i}$ I am aware that this makes perfect sense as an empirical estimator for the mean, however I am quite confused as to how about deducing this from the integral. We are integrating with respect to a differential of the ECDF, and deduce that the expression is $X_{i}$? I am struggling to even deduce what the result of $\int_{R}\mathbb{1}_{\{Xi \le x\}}$ evaluates to. Any help on the above would be much appreciated.
56591
This question is related to  , with the difference being that $F_X$ (the probability distribution function of random variable $X$) is an arbitrary continuous distribution function, not necessarily strictly increasing.  I think the proof is similar, but we have to take care of the possibility that $F_X$ may not be $1$-to-$1$.  I list my attempt below, and would appreciate it if someone can confirm if it's correct, and, in particular, if it can be improved.  Thanks a lot!  The goal is to show $F_Y(y)=y$ for any $y \in [0,1]$.  To do so, note that $F_Y(y)\triangleq\mathbb P(\{Y\le y\})$, and  $\{Y\le y\}=\{F_X(X)\le y\}=\{X\in F_X^{-1}([0, y])\}.$  Since $F_X$ is continuous, $F_X^{-1}([0,y])$ must be closed.  So it follows that   $\sup F_X^{-1}([0, y])=\max F_X^{-1}([0, y])=\max F_X^{-1}(\{y\}),$ which let's denote by $a$.    Therefore, $F_Y(y)=\mathbb P(\{X\le a\})=F_X(a)=y.\quad$  (Q.E.D.)
76786
For random variable $X$ that follows some distribution, $f(x)$ is the probability density function of that distribution if and only if $$\mathbb{E}[\phi(X)] = \int_{-\infty}^\infty \phi(x) f(x)dx$$ for all functions $\phi$.  Context: My professor used this in lecture to demonstrate a way to find the distribution of $cX$ given random variable $X$ that follows a specific distribution.   What is the name of this theorem? Have I missed any qualifications/caveats? I am in particular curious about whether "for all functions" is correct. My professor mentioned "positive, bounded, deterministic functions," but I am not sure what he meant by that. Where can I find a proof of this theorem?