Let $(P_t)_{t \geq 0}$ be a Poisson process with $\lambda &gt; 0$. I want to show that  $\mathbb{E}(P_s|P_s) = P_s$. I compute $$\mathbb{E}(P_s|P_s) = \sum_{x \in \text{Img}(P_s)} x \cdot \mathbb{P}(P_s = n|P_s = k) = \sum_{n=0}^{\infty} n \cdot \frac{\mathbb{P}(P_s = n \cap P_s = k)}{\mathbb{P}(P_s = k)}$$ Here I am stuck, since I don't basically don't know how to compute $$\mathbb{P}(P_s = n|P_s = k) = \frac{\mathbb{P}(P_s = n \cap P_s = k)}{\mathbb{P}(P_s = k)}$$
114688
My intuition told me E(X|X) is X. However,I get stuck when when I try to evaluate it from the definition. How do I define f(X|X) ? would it be constant? Thanks for your help.
34837
Being $X$ and $Y$ independent random variables distributed as $\mathrm{Exponential}(1)$ and let $T=X+Y$ Calculate $E(X|T)$.  This is my attempt at solving this:  Being $Z=\sum_{i=1}^{n}P_{i}$ where $P_{i}$~$\mathrm{Exp}(\lambda)$ then using it's MFG you have that $$M_{P_i}(t)=\frac{\lambda}{\lambda-t}$$ then $$M_{Z}(t)=M_{\sum{P_{i}}}(t)=(M_{P_{1}})...(M_{P_{n}})=(\frac{\lambda}{\lambda-t})^n$$ with this being the MFG of a $\mathrm{Gamma}$ distribution with parameters $(n,\lambda)$  With this result is easy to see that $T$~$\mathrm{Gamma}(2,1)$  Now $E[X|T]=\int xf_{X|T}(x|t)dx$  Then using Bayes' theorem we can make $$f_{X|T}(x|t)=\frac{f_{T|X}(t|x)f_{X}(x)}{f_{T}(t)}$$ then $$f_{T|X}(t|x)=Pr(T=t|X=x)$$ $$=Pr(X+Y=t|X=x)$$ $$=Pr(x+Y=t|X=x)$$ $$=Pr(Y=t-x)=f_{Y}(t-x)$$  Using this result we can now rearrenge $$f_{X|T}(x|t)=\frac{f_{Y}(t-x)f_{X}(x)}{f_{T}(t)}$$ $$=\frac{e^{-(t-x)}.e^{-x}}{\frac{te^{-t}}{\Gamma(2)}}$$ $$=\frac{1}{t}$$ then $$E[X|T]=\int (x)(\frac{1}{t})dx$$  And this where I got stuck because $1_{x\epsilon(0,\infty})$, I made this explanation step-by-step hoping that someone can se easily check where I went wrong. I would aprecciate a step-by-step answer, thanks.
91196
I am currently studying Poisson distribution.   I could understand \begin{align} \mathrm{E}(X) &amp;= \sum_{k=0}^{\infty} \frac{k \lambda^k e^{-\lambda}}{k!}= \\ &amp;= e^{-\lambda} \sum_{k=0}^{\infty} \frac{\lambda^{k+1}}{k!}=\\ &amp;= \lambda \end{align} but $\text{E}(X(X-1)) = \lambda^2$ why is this true ?   How can I check it ? I have no idea to check $E(X-1)$ and $E(X(X-1))$ instead of just $X$.
64287
Suppose $x_1$ and $x_2$ are independent Poisson random variables with parameters equal to $\lambda_1$ and $\lambda_2$ respectively. Show the sum of $x_1$ and $x_2$ is also a Poisson random variable using the moment generating function.   I know the function for Poisson is $\frac{\lambda^x e^{-\lambda}}{x!}$ and moment for it is $M_x(t) = e^{\lambda(e^t - 1)}$.  So I have to show $M_{x_1 + x_2}(t)$ is of the form $e^{k(e^t - 1)}$. But I don't know where to plug in $\frac{\lambda^{x_1} e^{-\lambda}}{x_1!} + \frac{\lambda^{x_2} e^{-\lambda}}{x_2!}$ into the moment function.
38629
Let $\{B_t,\mathcal{F}_t,t \geq 0\}$ be a standard one dimensional Brownian motion. Give an example of a random time $S$ with $P[0\leq S&lt; \infty]=1$ such that with $W_t:=B_{S+t}-B_S$, the process $W=\{W_t, \mathcal{F}^w_t;t\geq0\}$ is not a brownian motion.  How could I go about constructing such an example. I have no idea of how to proceed. I tried to invent an almost sure finite random time using the law of large numbers for Brownian motion $T_{\epsilon}=\{\sup_{s \geq0}: B_s\geq s\epsilon\}$ but I am not sure on how to handle the object $B_{T_{\epsilon}+t}-B_{T_{\epsilon}}$.  Any help would be appreciated
57271
Consider a Poisson process whose conditional intensity is  $$\lambda(t) = \alpha e^{-t}$$   starting at time $t=0$ for some parameter $\alpha&gt;0$.  I would like to simulate arrival/event/failure times as efficiently as possible. (I am not sure what the standard term is for the times you get from a Poisson process.) Is there a fast way to do this?
93796
Let $A$ and $B$ be two events of a probability space. Prove that  $\displaystyle|P(A\cap B)-P(A)P(B)|\leq \frac{1}4$   I think it's a very challenging problem, and I've made no progress so far ...  Can someone give me a hint ?
102499
Let $X$ be a discrete-time Markov process on some measurable space $(\mathscr X,\mathscr B)$. Let $B\in\mathscr B$ and  $$ \tau_B:=\inf\{n\geq 0:X_n\in B\} $$ is the first hitting time of $B$.    Let us assume that for any $x\in \mathscr X$  $$ \mathsf P_x\{\tau_B&lt;\infty\} = 1. $$ Can there exist $x$ such that $\mathsf E_{x}[\tau_B] = \infty$? Let us assume that for some $x&#39;\in \mathscr X$  $$ \mathsf P_{x&#39;}\{\tau_B&lt;\infty\} = 1. $$ Can it be that $\mathsf E_{x&#39;}[\tau_B] = \infty$?   No assumptions on the irreducibility (or any other properties) are imposed. I've tried to find a counterexample but didn't manage to do it.
107965
I am working on a problem with the following set:  $$S =  \{\varnothing,\{\varnothing\}\}$$  My solution was: $P(S) = \{\varnothing, \{\varnothing\},\{ \varnothing , \{ \varnothing \}\}\}$, but the solution on the textbook shows: $P(S) = \{ \varnothing , \{ \varnothing \}, \{\{\varnothing\}\}, \{ \varnothing , \{ \varnothing \}\}$  Where does the $\{\{\varnothing\}\}$ come from?  Thanks
72596
Given $n$ independent geometric random variables $X_n$, each with probability parameter $p$ (and thus expectation $E\left(X_n\right) = \frac{1}{p}$), what is $$E_n = E\left(\max_{i \in 1 .. n}X_n\right)$$    If we instead look at a continuous-time analogue, e.g. exponential random variables $Y_n$ with rate parameter $\lambda$, this is simple: $$E\left(\max_{i \in 1 .. n}Y_n\right) = \sum_{i=1}^n\frac{1}{i\lambda}$$  (I think this is right... that's the time for the first plus the time for the second plus ... plus the time for the last.)  However, I can't find something similarly nice for the discrete-time case.    What I have done is to construct a Markov chain modelling the number of the $X_n$ that haven't yet "hit". (i.e. at each time interval, perform a binomial trial on the number of $X_n$ remaining to see which "hit", and then move to the number that didn't "hit".) This gives $$E_n = 1 + \sum_{i=0}^n \left(\begin{matrix}n\\i\end{matrix}\right)p^{n-i}(1-p)^iE_i$$ which gives the correct answer, but is a nightmare of recursion to calculate. I'm hoping for something in a shorter form.
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.
119914
The theorem to prove is: $X$ is a positive continuous random variable with the memoryless property, then $X \sim Expo(\lambda)$ for some $\lambda$. The proof is explained in this , but I will type it out here as well. I would like to get some clarification on certain parts of this proof.  Proof Let $F$ be the CDF of $X$, and let $G(x)=P(X&gt;x)=1-F(x)$. The memoryless property says $G(s+t)=G(s)G(t)$, we want to show that only the exponential will satisfy this. Try $s=t$, this gives us $G(2t)=G(t)^2,G(3t)=G(t)^3,...,G(kt)=G(t)^k$. Similarly, from the above we see that $G(\frac{t}{2})=G(t)^\frac{t}{2},...,G(\frac{t}{k})=G(t)^{\frac{1}{k}}$. Combining the two, we get $G(\frac{m}{n}t)=G(t)^\frac{m}{n}$ where $\frac{m}{n}$ is a rational number. Now, if we take the limit of rational numbers, we get real numbers. Thus, $G(xt)=G(t)^x$ for all real $x&gt;0$. If we let $t=1$, we see that $G(x)=G(1)^x$ and this looks like the exponential. Thus, $G(1)^x=e^{xlnG(1)}$, and since $0 &lt;G(1) \leq  1$, we can let $lnG(1)=-\lambda$. Therefore $e^{xlnG(1)}=e^{-\lambda x}$ and only exponential can be memoryless.  So there are several parts that I am confused about:  Why do we use $G(x)=1-F(x)$ instead of just $F(x)$? What does the professor mean when he says that you can get real numbers by taking the limit of rational numbers. That is, how did he get from the rational numbers $\frac{m}{n}$ to the real numbers $x$? In the video, he just says that $G(x)=G(1)^x$ looks like an exponential and thus, $G(x)=G(1)^x=e^{xlnG(1)}$. How did he know that this is an exponential?
84959
A pitcher contains $X + 1$ blue balls and $Y + 1$ red balls. It is known that $X, Y$ are independent random variables, and it is given that $X \sim \mathrm{Poisson}(n), Y \sim\mathrm{Poisson}(2n)$. Calculate the limit: $\displaystyle\lim_{n\to \infty}\mathbb P (X + n = Y) $   I tried to condition on the value of $Y$ and to use the Law of total probability but it did not lead to a solution. ( I tried to sum: $ \sum_{i=0}^\infty\mathbb P(X+n=Y|Y=i)\mathbb P(Y=i) $ and get an expression that I can calculate its limit as n goes to infinity)
88460
How do I prove the following? I don't know where to start.     If $X$ is a random variable with $E^{\mathbb P}[X] = \mathbb P(X&gt;0)=1$   and $ \mathbb Q$ is the  probability measure defined by $ \mathbb Q(A)=E^{\mathbb Q}[X1_A] $ then $E^{\mathbb Q}[Y]=E^{\mathbb P}[XY]$
24263
I started with   $$P((A \cap B\mid C)=\frac{P((A \cap B)\cap C)}{P(C)}=\frac{P((A \cap C)\cap (B \cap C))}{P(C)}$$  by applying the basic property of conditional probability.  Then, I get stuck because I do not know how to come out with $P((A\mid B\cap C)$ and $P(B\mid C)$ from the above expression. I cannot think of any other possible multiplication rules that I can use. Please help!
50567
Well I attempted to try this but I failed to solve it: So $$\Phi_{p^{n}}(x)= \frac{x^{p^{n}} - 1}{\Phi_{1}(x) \Phi_{p^{2}}(x) \dots \Phi_{p^{n-1}}(x)}$$  Now I'm just stuck here. I saw a result somewhere on the net that said $\Phi_{p^n}(x) = \Phi_p(x^{p^{n-1}})$. I would just use that, but I don't see why that result is true or know how to show that's true.  How do I approach this?
38822
The questions asks to $ E[X_1|S_n]$ where $ S_n = \sum_{[n]} X_i $ with $X_i$ i.i.d. of finite expectation. My attempt was to consider an arbitrary Borel set, pull it back under $ S_n $ to get a set in $\sigma(S_n)$, and then average $X_1$ over that pull back and normalize:  $$ E[X_1|S_n] = E[X_1 1_{S_n^{-1}(B)}]/P[S^{-1}_n(B)]$$  for any Borel set $B$.  However, it just doesn't feel right - it seems just symbol pushing, without giving any intuition to what's happening.   Can anyone explain if this is correct, of if I missing something deep?