Can someone construct an example of this?  i.e., $E[X^2] &lt; \infty$ but $E[X^3] = \infty$. Results could be in terms of pdf, or cdf, or survival function. Justification would be appreciated
121868
Suppose $X$ is a random real variable with zero mean and finite second moment $\langle X^2\rangle$. Under what conditions can we give a bound (upper/lower) for the third moment $\langle X^3\rangle$?
45090
I need to find the following:  $$\lim_{x\to\infty} f(x)= \lim_{x \to \infty}\left (\frac{x}{x+1} \right )^x$$  I know that this limit = $\frac{1}{e}$ from plugging it into a calculator, but I have to prove it without using the fact that:  $$\lim_{x\to\infty} f(x)=\lim_{x \to \infty}\left (\frac{x}{x+k} \right )^x=\frac{1}{e^k}$$  I started by exponentiating:  $$\lim_{x\to\infty} f(x)=\lim_{x \to \infty}e^{\ln \left (\frac{x}{x+1} \right )^x}$$  and from here I've dropped the exponent $x$ in front of the $\ln$, and from here I'm getting stuck. Should I separate the $\ln$ like this?  $$\lim_{x\to\infty} f(x)=\lim_{x \to \infty}e^{x(\ln(x)-\ln(x+1))}$$  This doesn't seem to be leading me down the right path, but I'm not sure how else to do it. Is there a way to apply L'Hopital? If so, how?
83571
Let $(e_n)$ (where $ e_n $ has a 1 in the $n$-th place and zeros otherwise) be unit  standard vectors of $\ell_\infty$.   Why is $(e_n)$ not a basis for $\ell_\infty$?  Thanks.
51963
Evaluate the limit: $$\mathop {\lim }\limits_{x \to \infty } x\left( {{{\left( {1 + {1 \over x}} \right)}^x} - e} \right)$$  My attempts didn't yield a result. I'd be glad for a guidance. Thanks!
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 ?
82820
Kolmogorov proved, that, as one considers independent (not necessary equally distributed) Random Variables: $\{X_n\}_{n\ge0}\subseteq \mathcal L^2$   With $\mathrm{Var} (X_n)=\sigma^2_n$ and without loss of generality $E[X_n]=0$.  If $\sum_{n=0}^\infty \frac{\sigma^2_n}{n^2} \lt \infty$ then SLLN holds, that is:  $$\frac1n\sum_{k=0}^n X_k \rightarrow 0 \,\,\,\,\,\,\,\,\,\,\,\,\,\text{a.s.}$$  Something different:  Also it is known, that if $\sup_{n\ge0}\sigma^2_n =: v \lt \infty$ pairwise uncorrelated $(X_n)_{n\ge0}$ will suffice for SLLN to hold.  My question in view of this situation: Is it possible to weaken the condition of $(X_n)_{n\ge0}$ beeing independent and instead merely consider pairwise uncorrelated $(X_n)_{n\ge0}$?
27509
There is a pattern excercise:  First pattern: We have $X$ and the probability density function is usually:  $$ f(x) = \begin{cases}   {ax+bx^2} &amp;  {0\lt x \lt 1} \\   0 &amp; \text{else}   \end{cases} $$  Then we are given $E|X| = 0.6$   And usually asks to find:  a) $P\{X&gt;0.5\}$  b) $Var(X)$    Second pattern:  $$ f(x) = \begin{cases}   {cx^4} &amp;  {0\lt x \lt 2} \\   0 &amp; \text{else}   \end{cases} $$  Where c is a constant. Find:  a) Mean value of $X$  b) Variance of $X$     Third pattern (it's more like a reverse problem):  We are given $Y= \sqrt{X}$, find it's cumulative distribution function (CDF) and  it's probability density function (pdf).    Update:  The third one is transferred to a .  This is not homework. I'm self-studying and in need of some help. Thank you.
28863
If U has a $\chi^2$ distribution with v df, find E(U) and V(U).  By definition, $E(U) =\int^{\infty}_{0} u\frac{1}{\gamma(\frac{v}{2})2^\frac{v}{2}}u^{\frac{v}{2}-1} e^\frac{-u}{2}\,du  =\int^{\infty}_{0} \frac{1}{\gamma(\frac{v}{2})2^\frac{v}{2}}u^\frac{v}{2} e^\frac{-u}{2}\,du$.  How do I integrate this?  Note: This isn't a homework problem.
34581
I came across this question as an exercise, had a brief idea, but didn't know how to proceed.     Let $X \sim N(0, 1)$. What is the p.d.f of $|X|$ ?   I know the final p.d.f looks just like the right half of the original pdf, but extended vertically for a factor of 2.  Could someone please show mathematically what the p.d.f of variable |X| is going to be, and explain briefly? Thank you.
51037
Could someone point out a good source or provide justification as to whether the limit of a multivariate probability density function, as one or more or all variables tends to positive or negative infinity, goes to zero.  $$\lim_{x_{1}\to\pm\infty}f(x_{1},x_{2},...,x_{n})=0$$  $$\lim_{x_{1},x_{2}\to\pm\infty}f(x_{1},x_{2},...,x_{n})=0$$  $$\lim_{x_{1},x_{2},...,x_{n}\to\pm\infty}f(x_{1},x_{2},...,x_{n})=0$$  I have found this related question for the univariate case:
59359
Does $\displaystyle\lim_{n\to \infty}\ \frac{1+\sqrt[2]{2}+\sqrt[3]{3}+\cdots+\sqrt[n]{n}}{n}=1$?   Thanks a lot for your time and help.
52215
An insurance company's monthly claims are modeled by a continuous, positive random variable $X$ whose probability density function is proportional to $(x+1)^{-4}$ where $0 \lt x \lt \infty$   I need help understanding the reasoning behind the answer:  the PDF is given by:  $1 =\int_0^{\infty} \frac{k}{(1+x)^{-4}}$   Here is my first issue in the first step, where does K come from? We are given the pdf to be $\frac{1}{(x+1)^{4}}$ So to me it would stand that the function we are given is the PDF because it says thats the pdf, and we are then to find the expected value from here which would be equivalent to:  $\int_0^{\infty} x(1+x)^{-4}$ by definition of expected value  My guess would be that the words "proportional to" are key as to why we have a K introduced can someone explain this?  I understand everything except when taking the integral because and solving for 1 because I arrive at $\frac{3k}{(1+x)^4}$ rather than $\frac{-1}{3(1+x)^4}$  Taking the integral of $(1+x)^4$:  $u=1+x$ $du=1dx$ and   $\frac{u^{-4+1}}{-4+1} dx$  which gives $\frac{-3k}{(1+x)^3}$  Which gives me $k=\frac{1}{3}$ when solving $1=\frac{-3k}{(1+x)^4} \Big\vert_0^{\infty}$ rather than $k=3$ as the answer which is the correct answer. Can someone please explain this to me?
38967
Suppose $1&lt;p_1&lt;p_2&lt;\infty$, then show $L^{p_1}[a,b] \supset L^{p_2}[a,b]$.  I was able to show $\|f\|_{p_1} \le \|f\|_{p_2} (b-a)^{1/p_1 - 1/p_2}$ but I'm not sure how to proceed from here.
16099
I'm sure that this is a very basic question, but it has been bothering me for a while:  If $e=\lim\limits_{x\to \infty} (1+x^{-1})^x$, shouldn't $e=1$?  If $x$ is tending towards infinity, why wouldn't $x^{-1}$ tend to $0$ and $e=1^\infty=1$?  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?
34602
How do you show, that the expected value of a hyper-geometric random variable X with parameters $r$,$w$, and $n$ (a box contains $r$ red balls, $w$ white balls and $n$ balls are drawn without replacement) is $$\mathbf{E}(X)=\frac{rn}{r+w}?$$  I was able to reduce this to $$ \sum _{k=1}^{\infty}{r\choose k}{w\choose n-k}/{r+w\choose n} =\frac{n!(r+w-n)!r!w!}{(r+w)!k!} \sum _{k=1}^{\infty} \frac{1}{k!(r-k)!(n-k)!(w-n+k)!}. $$  No text I have seen yet proves this (and I think I know why mine doesn't either!). Any suggestions or references??
116368
I don't know how to evaluate it. I know there is one method using the gamma function. BUT I want to know the solution using a calculus method like polar coordinates.  $$\int_{-\infty}^\infty x^2 e^{-x^2}\mathrm dx$$  I will wait for a solution. Thank you.