Let $V,T$ be two random variables with supports $\mathcal{V},\mathcal{T}$, respectively. Let    $P_{V|T}$ denote the probability distribution of $V$ coditional on $T$ $P_{V,T}$ denote the probability distribution of $(V,T)$ $P_{T}$ denote the probability distribution of $T$, $P_{V}$ denote the probability distribution of $V$ $P_{T|V}$ denote the probability distribution of $T$ coditional on $V$   Suppose that $\mathcal{V}$ and $\mathcal{T}$ are finite. Then,  $$ E(V|T=t)= \sum_{v\in \mathcal{V}} v P_{V|T}(v|t)= \sum_{v\in \mathcal{V}} v \frac{P_{V,T}(v,t)}{P_{T}(t)}=\sum_{v\in \mathcal{V}} v \frac{P_{T|V}(t|v)P_V(v)}{P_{T}(t)} $$ Moreover,  $$ \underbrace{E(V|T=t)\geq 0 \leftrightarrow \sum_{v\in \mathcal{V}} v P_{T|V}(t|v)P_V(v)\geq 0}_{(*)} $$ Now, I want to write condition $(*)$ (and in particular the right hand side)  when $V$ and $T$ are continuous random variables, without using probability density functions but only probability measures or cumulative distribution functions. Could you help?    My attempt: I managed to write $(*)$ when $\mathcal{T}$ is finite and $V$ is a continuous random variable: $$ E(V|T=t)\geq 0 \leftrightarrow \int_{\mathcal{V}} v P_{T|V}(t|v)dP_V(v)\geq 0 $$ But I'm struggling to extend to the case when also $T$ is a continuous random variable.
123179
I'm looking for some correct notation. Consider the random variable $V$ with support $\mathcal{V}$ and probability distribution $P_V$. Consider a function $u:\mathcal{V}\rightarrow \mathbb{R}$.   Let $\mathcal{V}$ be finite and suppose I want to  compute the expected value of $u(V)$. This is   $$ \sum_{v\in \mathcal{V}} u(v)\times  P_{V}(v) $$  Suppose now instead that $P_V$ is continuous. What is the correct notation to indicate the formula of the expected value of $u(V)$ without passing through the definition of density? I though about using $$ \int_{\mathcal{V}} u(v)\text{ } dP_{V}(v) $$  Is this correct?
12665
Let $X_1,X_2,\ldots,X_n$ be i.i.d variables having a Pareto distribution with density $$f(x)=\frac{a\theta^a}{x^{a+1}}1_{x&gt;\theta}\,,$$ where $a,\theta&gt;0$. What is the distribution of $\sum\limits_{j=1}^n \ln\left(\frac{X_{(j)}}{X_{(1)}}\right)$ ?   Suppose $\mathsf{Gamma}(p,\alpha)$ denotes the density $g(t)\propto e^{-\alpha t}t^{p-1}1_{t&gt;0}$.  We have $$T=\sum_{j=1}^n\ln\left(\frac{X_{(j)}}{X_{(1)}}\right)=\sum_{j=1}^n\ln(X_{(j)})-n\ln(X_{(1)})=\sum_{j=1}^n \ln X_j-n\ln X_{(1)}$$  Now,   \begin{align}&amp;\ln(X_j/\theta)\stackrel{\text{i.i.d}}{\sim}\mathsf{Exp}\text{ with mean }1/a\qquad,\,j=1,\ldots,n \\&amp;\implies\sum_{j=1}^n\ln(X_j/\theta)=\sum_{j=1}^n\ln X_j-n\ln \theta\sim\mathsf{Gamma}(n,a) \end{align}  I could show that $X_{(1)}$ has another Pareto density, so that $$\ln\left(\frac{X_{(1)}}{\theta}\right)=\ln X_{(1)}-\ln \theta \sim \mathsf{Exp}\text{ with mean }1/(na)$$   Not sure if the last two facts help me get the exact distribution of $T$.    Edit:  Turns out this was rather simple had I simply rewritten $T$ as   $$T=\sum_{j=1}^n \ln\left(\frac{X_j}{X_{(1)}}\right)=\sum_{j=1}^n(\ln X_j-\ln X_{(1)})=\sum_{j=1}^n (Y_j-Y_{(1)})\,,$$  where $Y_j=\ln (X_j/\theta)$. Since $aY_j\sim\mathsf{Exp}(1)$, using  result I have $a T\sim \mathsf{Gamma}(n-1,1)$.   This is equivalent to $T\sim \mathsf{Gamma}(n-1,a)$ or $2aT\sim \chi^2_{2n-2}$.
71004
I'm working on a problem from Casella and Berger's Statistical Inference. X is distributed as Poisson$(\theta)$ and Y is distributed as Poisson$(\lambda)$, with X and Y being independent. We let U = X + Y and V = Y, and the conditional pdf of V|U is:   $\hspace{15mm}\large f(v|u) = \large \frac{f(u,v)}{f(u)} = \huge \frac{\frac{\theta^{u-v}}{(u-v!)}*\frac{\lambda^ve^{-\lambda}}{v!}}{\frac{(\theta+\lambda)^ue^{-(\theta+\lambda)}}{u!}}$  Apparently this simplifies to   $\hspace{15mm}\Large {u \choose v} (\frac{\lambda}{\theta+\lambda})^v(\frac{\theta}{\theta+\lambda})^{u-v}$  I don't see how the $\Large(\frac{\lambda}{\theta+\lambda})^v$ is possible. I keep ending up with $\Large(\frac{\lambda}{\theta})^v$ when I simplify the problem.
40744
Let $X$ be a continuous random variable having uniform distribution on $[0, 2\pi]$. What distribution has the random variable $Y=\sin X$ ? I think, it is also uniform. Am I right?
44211
Let $X$ be a random variable with a continuous and strictly increasing c.d.f. function $F$ (so that the quantile function $F^{−1}$ is well-deﬁned). Deﬁne a new random variable $Y$ by $Y = F(X)$. Show that $Y$ has a uniform distribution on the interval $[0, 1]$.  My initial thought is that $Y$ is distributed on the interval $[0,1]$ because this is the range of $F$.  But how do you show that it is uniform?
91434
When I studied Bayesian statistics, a question about the notation of Bayes' Theorem came to my mind. Below is the density function version of Bayes' Theorem, where $y$ is data vector and $\theta$ is the parameter vector:  $$ p(\theta|y)=\frac{p(y|\theta)p(\theta)}{p(y)} $$ The numerator on the right handside can be written as: $$ p(y,\theta) $$ which is the joint probability distribution of $y$ and $\theta$, then Bayes' theorem could be written as: $$ p(\theta|y)=\frac{p(y,\theta)}{p(y)} $$ Furthermore, $$  p(\theta|y)\propto p(y,\theta) $$ Am I right on this? I think it does not look right. Because the posterior is the proportional to the joint density function. But where is the mistake?
112403
I was wondering how to (or if it is even possible) find the continuous joint distribution between two random variables $x$ and $y$ when you know the continuous marginal density distributions of both $x$ and $y$, and we know there is a correlation between the two random variables, i.e., ${\rm Cor}(x,y)$ = $\rho$ which is of unknown magnitude?
39962
I would like to show that:  $\hspace{2mm} E[X|X&lt;x] \hspace{2mm} \leq  \hspace{2mm} E[X] \hspace{2mm} $ for any $x$   X is a continuous R.V. and admits a pdf. I'm guessing this isn't too hard but I can't come up with a rigorous proof. Thanks so much.
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?
18879
Let $U_1, \cdots, U_n$ be $n$ i.i.d continuous uniform random variables on $(0,1)$ and their order statistics be $U_{(1)}, \cdots, U_{(n)}$.  Define $D_i=U_{(i)}−U_{(i−1)}$ for $i=1, \cdots, n$ with $U_0=0$.  I am trying to figure out the distribution $\lim_{n\to\infty} \#\{i = 1, \cdots, n \mid nD_i \geq z\}/n$. This must be very well known, but I cannot find a reference.
21382
In the book on Brownian Motion by Yuval Peres, he makes the following claim:     Suppose that $\{ B(t); \ t \geq 0 \}$ is a Brownian Motion and $U$ is an independent random variable, uniformly distributed on $[0, 1]$. Then the process $\{ \tilde{B} (t); \ t \geq 0 \}$ defined as,   $$\tilde{B}(t) = \begin{cases} B(t)  &amp; \mbox{ if $t \neq U$} \\ 0  &amp; \mbox{ if $t = U$} \end{cases}$$      has the same finite-dimensional distributions as a Brownian motion, but is discontinuous if $B(U) \neq 0$, i.e. with probability one, and hence this process is not a Brownian motion.   I'm having difficulty in proving that the finite dimensional joint distribution ( basically the joint distribution of $( \ \tilde{B}(t_1), \ \tilde{B}(t_2), \tilde{B}(t_3), \dots \tilde{B}(t_n) \ )$ ) is the same as the finite dimensional distribution of $B(t)$. Can you please give me some clues.
70467
$(\Omega, \mathcal{F}, P)$ is a probability space. $X$ is a r.v. defined on it, and $\mathcal{G}_1, \mathcal{G}_2$ are two $\sigma$-algebra, can we claim the following: $$ \mathbb{E}\{\mathbb{E}[X|\mathcal{G}_1]|\mathcal{G}_2\}=\mathbb{E}\{\mathbb{E}[X|\mathcal{G}_2]|\mathcal{G}_1\}=\mathbb{E}[X|\mathcal{G}_1\cap\mathcal{G}_2]. $$ If this is not true, why the following is true: $X_1,X_2,\ldots,X_n$ are independent r.v. on the same probability space, $Z=f(X_1,X_2,\ldots,X_n)$ is another r.v. And we have: $$ \mathbb{E}\{\mathbb{E}[Z|X_1,\ldots,X_{i-1},X_{i+1},\ldots,X_n]|X_1,\ldots,X_i\}=\mathbb{E}[Z|X_1,\ldots,X_{i-1}]. $$
102954
Let $X$ ~ $U(0,2)$ and $Y$ ~ $U(-10,10)$ be two independent random variables with the given distributions. What is the distribution of $V=XY$?   I have tried convolution, knowing that   $$h(v) = \int_{y=-\infty}^{y=+\infty}\frac{1}{y}f_Y(y) f_X\left (\frac{v}{y} \right ) dy$$  We also know that $f_Y(y) = \frac{1}{20}$,    $$h(v)= \frac{1}{20} \int_{y=-10}^{y=10} \frac{1}{y}\cdot \frac{1}{2}dy$$ $$h(v)=\frac{1}{40}\int_{y=-10}^{y=10} \frac{1}{y}dy$$  Something tells me, there is something weird here since it is discontinuous at 0. Please help.
11459
$X \sim \mathcal{P}( \lambda) $ and $Y \sim \mathcal{P}( \mu)$ meaning that $X$ and $Y$ are Poisson distributions. What is the probability distribution law of $X + Y$. I know it is $X+Y \sim \mathcal{P}( \lambda + \mu)$ but I don't understand how to derive it.
23086
Let $u$ be a distribution on $\mathbb{R}^n$ with support = $\left\{0\right\}$. Then there exists $N$ such that $u$ has order $N$. Let $\chi\in C_0^{\infty}(\mathbb{R}^n)$ a smooth function with  $\chi(x) = 1$ for $0\leq |x|\leq 1 $,   $\chi(x)\in [0,1]$ for $1\leq |x|\leq 2$,  $\chi(x)=0$ for $|x|\geq 2$  Denote $ \chi(x/r) = k_r(x)$ for $r\in (0,1]$. For the case $N=0$ I want to show that there exists $c_1$ such that $$|\left\langle u,\phi \right\rangle|\leq c_1 |\phi(0)|  $$ for all $\phi \in C_0^{\infty}(\mathbb{R}^n)$. The idea would be to apply $u$ on $\phi = k_r\phi+(1-k_r)\phi$ and letting $r\to 0$.   So then I guess we can write $$ |\left\langle u,\phi \right\rangle| \leq |\left\langle u,k_r\phi \right\rangle| + |\left\langle u,(1-k_r)\phi \right\rangle |$$ But not really sure how this follows...  Moreover how can i show that $\left\langle u,\phi \right\rangle = \left\langle u,\phi \chi\right\rangle = \left\langle u,\chi\right\rangle\phi(0) $?
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.
97524
Let $T_n$ be a random variable with $T_n=X_1+...+X_n$ where the $X_i$'s are iid. Further we set $N(t)=max\{ n: T_n\leq n\}$ with the property $\Pr(N(t)&lt;\infty)=1$.  I want to prove that $\mathbb{E}[N(t)]&lt;\infty$.   I have done this so far. Let $F_n$ be the distribution function of $T_n$. $$\mathbb{E}[N(t)]=\sum_{n=1}^{\infty} n P(N(t)=n)=\sum_{n=1}^{\infty} n (F_n(t)-F_{n+1}(t))=\sum_{n=1}^{\infty} F_n(t)$$  Further I know that $0 \leq F_n \leq 1$. But does this implies $\mathbb{E}[N(t)]&lt;\infty$ ? I don't think so... I don't know what I can do further...   Can anybody help me how I can proof it?   Thank you for any help!!!