Show that $E[x]=\int_0^\infty(1-F_X(x))dx$ The solution is as follows: $$\begin{align} \int_0^\infty P(X&gt;x)dx &amp;=\int_0^\infty \int_x^\infty f_X(y)dydx\\ &amp;=\int_0^\infty  \int_0^y dxf_X(y)dy\\ &amp;=\int_0^\infty yf_X(y)dy\\ &amp;=E[X] \end{align}$$  I don't understand step 3 in the solution, specifically while the limit of integration changed from $\infty$ to $y$.
121825
Let $X$ be a continuous non-negative random variable (i.e. $R_x$ has only non-negative values). Prove that $$E(X) = \int_{0}^{\infty} P(X&gt;x)\,dx  = \int_{0}^{\infty} (1-F_X(x))\,dx$$ where $F_X(x)$ is the CDF for $X$. Using this result, find $E(X)$ for an exponential ($\lambda$) random variable.   I know that by definition, $F_X(x) = P(X \leq x)$ and so $1 - F_X(x) = P(X&gt;x)$  The solution is: $$\int_{0}^{\infty} \int_{x}^{\infty} f(y)\,dy dx    = \int_{0}^{\infty} \int_{0}^{y} f(y)\,dy dx  = \int_{0}^{\infty} yf(y) dy.$$  I'm really confused as to where the double integral came from. I'm also rusty on multivariate calc, so I'm confused about the swapping of $x$ and $\infty$ to $0$ and $y$.  Any help would be greatly appreciated!
71393
Find a general formula for evaluating integrals of the form $\displaystyle\int_0^\infty x^a e^{-bx}dx.$   I've been pondering about this question for some time. Obviously integration by parts is way too inefficient. I know there's recursion involved and maybe there's a way to find a general formula through intuition.
108648
Suppose that $\int_{\mathbb R} |f(x)| dx &lt; \infty$.     Can we say that $\int^{\infty}_{n} |f(x)| dx \to 0$ as $n \to \infty$ If yes, how to justify?
83140
Given:  Let $X$ be exponential with parameter $\lambda$, that is   $$ f_X(x) = \begin{cases} \lambda e^{-\lambda x} &amp; \text{if }x&gt; 0, \\ 0 &amp;\text{for }x\leq 0. \end{cases} $$  where $\lambda&gt;0$ is called the rate of the distribution.  Question:  Show that this is a valid probability density function (pdf).  Attempt:  Need to show that  $$ f(x) = \int^{\infty}_{-\infty} \lambda e^{-\lambda x}\,  dx=1\\\\ $$ Thus, \begin{eqnarray*} f(x)&amp;=&amp;\int^0_{-\infty} \lambda e^{-\lambda x} \, dx+\int^\infty_0 \lambda e^{-\lambda x} \,  dx\\[8pt] &amp;=&amp;0+\int^\infty_0 \lambda e^{-\lambda x} \, dx \\[8pt] &amp;=&amp;\lambda\int^\infty_0 e^{-\lambda x} \, dx \\[8pt] &amp;=&amp;1 \end{eqnarray*}  The problem is that my professor just told me that I skip a step right before I derive $1$. Any insight?  EDIT:  Given the great feedback, here is the answer:  \begin{eqnarray*} f(x)&amp;=&amp;\int^{\infty}_{-\infty} \lambda e^{-\lambda x} dx=1\\\\ f(x)&amp;=&amp;\int^{\infty}_{-\infty} f_X(x)dx\\ &amp;=&amp;\int^{0}_{-\infty} 0\,\,\, dx+\int^{\infty}_{0} \lambda e^{-\lambda x} dx\\ &amp;=&amp;0+\int^{\infty}_{0} \lambda e^{-\lambda x} dx\\ &amp;=&amp;\int^{\infty}_{0} \lambda e^{-\lambda x} dx\\ &amp;=&amp;\int^{\infty}_{0} e^{-\lambda x} \Big(\lambda \,\,\,dx\Big)\\ &amp;=&amp;\int^\infty_0e^{-u} du\\ &amp;=&amp;-e^{-u}\Big|^{\infty}_0\\ &amp;=&amp;1-\lim _{x \to \infty} e^{-u}\\ &amp;=&amp;1\\ \end{eqnarray*}
83864
In fact the real question is: Does $\lim\limits_{n\to\infty}1^{n}=e$?.  I know that   $$ \lim\limits_{n\to\infty}\left(1+\dfrac{1}{n}\right)^n=e, $$  So, can we say that $1^\infty=e$?  And, by logic, This product  $$\underset{\text{infinity times}}{1\cdot1\cdot\ldots\cdot1},$$ gives $1$ not other value.
95551
I'd like some help with the following integral:  $$\int_0^\infty \lfloor x \rfloor e^{-x}\mathrm dx .$$  Thanks.
44935
I ask this because I'm trying to understand a proof of the expected value of a non negative random variable is equal to $\int_{0}^{\infty}(1-F(x))dx$ ($F$ is the distribution function of $X$) when $E[X]$ exists, and in some step I have $\lim\limits_{x\to\infty}({x·P[X&gt;x])}=0$ and I don't know why it verifies. The problem I see is the indetermination in the limit $\infty·0$.
96066
What is $E|X-Y|$ when $X$, $Y$ are i.i.d and uniform on $[0,1]$?   I want to do something like   $$ E(|X-Y|)=\iint_{x&gt;y}(x-y)\;dxdy+\iint_{y&gt;x}(y-x)\;dxdy=2\iint_{x&gt;y} (x-y)\;dxdy $$  But am not sure how to solve that last double integral or if this setup is right.
114331
Let $X : \Omega \to \mathbb{N} $ be a r.v. Then $$ \mathbb{E} \{ X \} = \sum_{n=0}^{\infty} P(X &gt; n) $$ Attempt: $$ \mathbb{E} \{ X \} = \sum_j j P(X = j ) = \sum_j j(1 - P(X &gt; j) - P(X&lt;j ) )$$ But here it gets complicated. Am I on the right track?
45796
Let $f:(0,\infty) \to R$ be differentiable. Suppose that $\lim_{x\to\infty}(f(x)+f'(x))=L$. Show that $\lim_{x\to\infty} f(x) = L$ and $\lim_{x\to\infty} f'(x) = 0$. (Hint: Write $f(x) = e^xf(x)/e^x$ and use l’Hopital’s Rule.)    My working for $\lim_{x\to\infty}f'(x)=0$:  For $\lim_{x\to\infty}f'(x) = 0$, I let $f(x) = e^xf(x)/e^x$ and applied quotient rule which then cancels off $e^{2x}$ and I'm left with $\lim_{x\to\infty}(f(x)+f'(x)-f(x))$. Can I then equate this with $\lim_{x\to\infty} \left(f(x)+f'(x)\right) - \lim_{x\to\infty}f(x)$ which then gives $L-L=0$? Is this step correct?
109939
Let $f$ be a continuous function. I wish to show $$\lim_{n \rightarrow \infty} \int_0^1 n x^n f(x) \; dx = f(1)$$ I can try to split up the integral over intervals $[0, 1-\delta]$, $[1-\delta, 1]$. The integral over $[0, 1-\delta]$ vanishes as $n \rightarrow \infty$, but I am having difficulty estimating the integral over $[1-\delta, 1]$. Thanks!
102007
I'm given a joint pdf  $f_{X,Y}(x,y)=2e^{-x-y}, 0&lt;x&lt;y, 0&lt;y $  and asked to compute $P(Y&lt;3X)$. To do this, I let $Y=3X$ (the boundary) and found that the region of integration is under this line.  To find $P(Y&lt;3X)$, it seems to me that the integral for this region be written as   $P(Y&lt;3X)=\int_0^\infty \int_0^{3x} 2e^{-x-y} dy dx  $  However this isn't right, as the boundaries are actually   $P(Y&lt;3X)=\int_0^\infty \int_x^{3x} 2e^{-x-y} dy dx  $  Why is this? I'm having a hard time seeing how to construct these boundaries (especially the fact that the $y$ boundary goes from $x$ to $3x$ instead of $0$ to $3x$)  Thank you.
114326
Let $f$ be a continuous, nonnegative, real-valued function and $$\int_0^{\infty}f(x)dx&lt;\infty.$$ Prove that $$\lim_{n\to\infty}\frac1{n}\int_0^{n}xf(x)dx=0.$$  A start: If $\lim\limits_{n\to\infty}\int_0^{n}xf(x)dx$ is finite, then it's obvious. Otherwise, perform L'Hopital's rule, we get $\lim nf(n)$, which we want to show is $0$.
96658
How can it be shown that   $$\lim_{p\to\infty}I(p)= \lim_{p \to \infty}\int^{1}_0 (x^x)^{\scriptscriptstyle {(x^x)^{(x^x)^{(x^x)^{(x^x)^{(x^x)...(p \; times)}}}}}} dx= \frac{\pi^2}{12}$$      $I(1)=\int^{1}_0 x^x dx=\sum_{n=0} ^{\infty} \frac{1}{n!}\int^{1}_0 \ln(x)^nx^n dx$   also $\int^{1}_0 \ln(x)^nx^n dx=(-1)^n (1+n)^{-1-n}n!$ .  I don't know how to calculate $I(p)$ beyond $p=1.$  Note: This integral was proposed on Romanian Mathematical Magazine and two solutions can be found .
54941
$$ \int_0^{\infty } \frac{\log (x)}{e^x+1} \, dx = -\frac{1}{2} \log ^2(2) $$  Anyone an idea on how to prove this?
68926
I want to find the following integral: $$\int_{0}^{\infty} \frac{e^{3x}-e^x}{x(e^{3x}+1)(e^x+1)} dx$$ I have been trying to apply simple substitutions like $u=e^x$, to simplify the integral without any luck. I simplified it to  $$\int_{0}^{\infty} \frac{1}{x(e^x+1)}-\frac{1}{x(e^{3x}+1)}dx$$  which seems to have increased my problems.  Wolfram Alpha gives the answer to be 0.5493... and the graph looks like a Gaussian distribution.  Somehow, I feel that this integral is very non-trivial due to the 'troublesome' $x$ in the denominator.
118167
I have the following problem:  Suppose that $Y \sim \text{Poi}(1)$. Let $y \ge 0$ be an integer. Given that $Y = y$, suppose that $X$ has the discrete uniform law on $\{ 0, 1, \dots, y \}$. (a) Show that $E[X \vert Y] = \dfrac{1}{2}Y$, and use this to find $E[X]$.  The provided solution is as follows:  We are given that $P(X = x \vert Y = y) = (y + 1)^{-1}$ for $x \in \{ 0, 1, \dots, y \}$, and $P(X = x \vert Y = y) = 0$ otherwise. (a) From the conditional PMF, we can compute the conditional expectation of $X$ given $Y = y$: $$\begin{align} E[X \vert Y = y] &amp;= \sum_{x = 0}^y xP(X = x \vert Y = y) \\ &amp;= (y + 1)^{-1} \sum_{x = 0}^y x \\ &amp;= (y + 1)^{-1} \left[ \dfrac{y(y + 1)}{2} \right] \\ &amp;= \dfrac{y}{2}. \end{align}$$ Therefore, $E[X \vert Y] = \dfrac{Y}{2}$. Now, since $Y \sim \text{Poi}(1)$, we know that $E[Y] = 1$. We now compute $E[X]$ using the law of total expectation: $$\begin{align} E[X] &amp;= E[E[X \vert Y]] \\ &amp;= E \left[ \dfrac{Y}{2} \right] \\ &amp;= \dfrac{1}{2} E[Y] \\ &amp;= \left( \dfrac{1}{2} \right) (1) \\ &amp;= \dfrac{1}{2}. \end{align}$$  How is it that $\sum\limits_{x = 0}^y x = \dfrac{y(y + 1)}{2}$? I would appreciate it if someone would please take the time to explain this.