There are two independent RVs $X \sim \mathcal{U}(-1,5)$ and $Y \sim \mathcal U(-5,5)$.  Let $A = \{ X \ge Y \land Y \ge -1 \land Y \le 1\}$ and $B = \{ X \le 1\}$.  What is $\mathbb P (A | B)$?  My idea is to apply Bayes' rule, i.e.  $$ \mathbb P(A|B) = \frac{\mathbb P(B|A) \mathbb P(A)}{\mathbb P(B|A)\mathbb P(A) + \mathbb P(B|\lnot A) \mathbb P(\lnot A)}$$  I understand that this works nicely as long as $\mathbb P(B|A)$ is easier to compute.  However, I find that  $$ \mathbb P(B|A) = \frac{\mathbb P(A|B) \mathbb P(B)}{\mathbb P(A|B)\mathbb P(B) + \mathbb P(A|\lnot B) \mathbb P(\lnot B)}$$  ... which I feel does not bring me closer to the solution. The unconditional probabilities are fairly straight forward, but the conditional ones keep piling up.  How should I procede from there?
123303
Given two independent random variables $X \sim \mathcal{U}[-1,5]$ and $Y \sim \mathcal{U}[-5,5]$, what is $$\mathbb{E}\{Y - X | X \le 1, Y &gt; X, Y \in [-1,1] \}\,?$$  I managed to compute the respective probability   $$ \mathbb{P}\{ X \le 1, Y &gt; X, Y \in [-1,1] \} = \mathbb{P}(X \le 1) \mathbb{P}(Y &gt; X) \mathbb{P}(Y \in [-1,1]) = 1/3 * 1/2 * 1/5 = 1/30$$  How do I continue from there?
58873
Does this identity hold for all events?  $$ P(A|B) = 1-P(A'|B) $$  Logically speaking, if the probability of $A$ given $B$ occurred is $X$, shouldn't the probability that $A$ does not occur, $A'$, given $B$, be similarly $1-X$?  There is a related question .  This is the closest that I could get to proving (or disproving) it:   $P(A\cap B)=P(A)-P(A' \cap B)$ $P(A|B)P(B)=P(A)-P(A'|B)P(B)$ $\therefore P(A'|B)= \frac {P(A)} {P(B)}-P(A|B)$   Are there are certain formulae which can be used to prove this? Or does the identity only hold under certain situations, and if so, what kind of situations?  Thanks.
63896
Suppose that we have the situation as depicted in the figure: a random experiment which has 4 outcomes $x_1, ..., x_4$ and two events $A$ and $B$. Also assume that $P(x_i)=0.25$.  Now, since $P(A \cap B) = P(A)P(B) = 0.25$, by definition the events $A$ and $B$ are independent. This does not make sense to me. Why are they independent? Is there an intuitive explanation?  Another thing is, suppose that there is a 5th outcome, $x_5$, outside of $A \cup B$, in that case the events $A$ and $B$ are no longer independent since  $P(A \cap B) \ne P(A)P(B)$. This result also does not make sense to me.
72927
Let $\mathbb P(X=1) = \mathbb P(X=-1) = 1/2$. Define       $$X_n = \begin {cases} X &amp; \text{with probability } 1- \frac{1}{n}\\  e^n &amp; \text{with probability }  \frac{1}{n} \end {cases}$$      Does $X_n \leadsto X$ (i.e., converge in distribution)? Does $X_n \stackrel{P}{\rightarrow} X$? Does $\mathbb{E}[(X-X_n)^2] \to 0$?   I think $X_n \leadsto X$ because just by picturing the CDF  as n goes to infinity, their CDFs converge. Can I get some confirmation on this?  But for convergence in probability, I'm having trouble with the limits.   To do this, what I'm trying to do is find $\mathbb P(|X_n-X| \leq \epsilon)$ and then just take 1 minus that as shown on this site:   If $X_n = X$, then $|X_n - X|$ is just $0$ so for all $n$, $|X_n -X| &lt; \epsilon$ But if $X_n = e^n$ then I have the two cases for $X$. For $X=1$, $|X_n-X| = |e^n-1| \leq \epsilon$. and For $X=-1$, $|X_n-X| = |e^n +1| = e^n + 1 \leq \epsilon$. This is where I am stuck. I'm trying to find conditions relating $n$ and $\epsilon$ similiar to here the example on the site. But I'm not sure how to get that from $|e^n - 1| \leq \epsilon$ and  $e^n + 1 \leq \epsilon$ .  Am I even on the right track?
77131
I have a question. Let's suppose that the two random variables $X1$ and $X2$ follow two Uniform distributions that are independent but have different parameters:  $X1 \sim Uniform(l1, u1)$  $X2 \sim Uniform(l2,u2)$  If we define X3 as the maximum of X1, and X2, i.e., $X3 = max(X1, X2)$, what kind of distribution would it be? It is certainly not a uniform and I can calculate the cumulative probability, i.e., $p(X3 &lt;= a)$, because $p(X3 &lt;= a) = p(X1 &lt;= a) p(X2 &lt;= a)$. But, I wonder if there exists any density function that can model X3.
121387
Is it true or false that $\lvert P([0, 1]) \setminus P((0, 1)) \rvert &gt; \lvert P(\mathbb{N}) \rvert$ ?  I need a hint because I don't see a way to solve this
100720
Let $X \sim Bin(n,p)$   Part A.  Show that the argument      "Then $X/n \sim Bin()$ with $E[X/n]=p$and $Var[X/n]=pq/n$"    is false.   My book says we can prove this result using a moment generating function. But to me it makes no intuitive sense: We have Bin-distributed RV $X$ and if we rescale it with $1/n$ of course it will be Bin-distributed.  And we know from standard formulas that $E[aX]=aE[X]$ and $Var[aX] = a^2Var[X]$ so I think the argument is correct.   (For full disclosure, the argument we should falsify according to the book is "Let X be a binomial random variable with mean np and variance np(1−p). Show that the ratio X/n also has a binomial distribution with mean p and variance p(1−p)/n." But I think the argument is correct. )    Ok, now I get part A. Here is Part B. Prove it using the moment generating function of $X/n$.   Attempt: I get $$m_X(t)=E[\exp(tX/n)]=(m_X(t))^{1/n}$$
21271
I am unable to compute conditional probability(x|x>y) in the above question. Also, I am unable to determine the region of integration for calculation of the above expectation.
34282
I have random variables $X_{1}, X_{2}, \dots X_{n}$ that are each Unif$\{1, \dots, n\}$, and I want to calculate $$P(\{X_{1}, X_{2}, \dots X_{n}\}=\mathbb{N})$$  My intuition is that this means each $X_{n} = n$, which occurs with probability $\frac{1}{n!}$.  Does this seem correct?
38218
Let $X \sim Gamma(3,3)$ and $Y \sim Exp(1)$.  How do I calculate $P(X&gt;Y)$?  I believe I rewrite it as $P(X-Y&gt;0)$ but I am unsure how to calculate $X-Y$ for two different distributions?
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 ?
16379
Let $X$ and $Z$ be independent, with $X\sim N(0,1)$, and with $\textbf{P}(Z=1)=\textbf{P}(Z=-1)=\frac{1}{2}$. Let $Y=XZ$ (i.e., $Y$ is the product of $X$ and $Z$).   (a) Prove that $Y\sim N(0,1)$.  (b) Prove that $\textbf{P}(|X|=|Y|)=1$.   (c) Prove that $X$ and $Y$ are not independent.   (d) Prove that $\textbf{Cov}(X,Y)=0$.  (e) It is sometimes claimed that if $X$ and $Y$ are normally distributed random variables with $\textbf{Cov}(X,Y)=0$, then $X$ and $Y$ must be independent. Is that claim correct?   My Work: (a) Is really where I am a little stuck. In order to show that $Y\sim N(0,1)$, I want to show that they have the same distribution laws. Thus, I want to show that $\mathcal{L}(Y)=\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}\textbf{1}_B(t)\lambda(dt)$ given that $\mathcal{L}(X)=\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}\textbf{1}_B(t)\lambda(dt)$.   However, I don't know how to go about that. Another approach I have considered is to show that for any Borel measurable function $f:\mathbb{R}\to\mathbb{R}$,$\space$ $\textbf{E}(f(X))=\textbf{E}(f(Y))$ for which each is well defined. Which approach do you think is better and How should I start. Any help will be appreciated.
66921
I have a good intuition that $A$ is independent of $B$ if $P(A \vert B) = P(A)$, and I see how you can easily derive from this that it must hold that $P(A,B) = P(A)P(B)$.    But the first statement is not normally taken as a definition; instead the second is.    What is the intuition, or even derivation behind defining $A$ and $B$ as independent iff $P(A, B) = P(A)(B)$?    The kind of explanation I am looking for would be one similar to that given by Jaynes for the definition of conditional probability in the first chapter of Probability: The Logic of Science, or even a Kolmogorov axiomatic explanation would help.
46678
I found this question while browsing through "The Probabilistic Method", by Noga Elon.  Let X and Y be 2 independent and identically distributed real valued random variables. Prove that: $$P(|X-Y| \leq 2) \leq 3P(|X-Y| \leq 1)$$  So I tried the following:  $$P\{|X-Y| \leq 2\} = P\{|X-Y| \leq 1\} + P\{X-Y \in (1,2]\cup[-2,-1)\}$$ $$= P\{|X-Y| \leq 1\} + P\{X-Y \in (1,2]\} + P\{X-Y \in [-2,-1)\}$$ $$= P\{|X-Y| \leq 1\} + 2P\{X-Y \in (1,2]\}$$ where the last step follows because $X-Y$ has a symmetric distribution. NB: A random variable Z has symmetric distribution if  $$P(Z \leq z) = P(Z \geq -z) \quad \forall z \in \mathbb{R}$$  Thus the problem boils down to showing $$P\{X-Y \in (1,2]\} \leq P(|X-Y| \leq 1)$$ and I would be done. Unfortunately, I don't know how to proceed from here. I appreciate any help, hints, useful comments etc. I receive.
74362
If $x_1,x_2,x_3$ and $x_4$ are independent identically distributed  random variable with continuous distribution .what is the probability of $$P=p(x_1 &lt; x_2&lt; \max(x_3,x_4))$$ I know the exhaustive number of cases are $p(x_1&lt;x_2&lt;x_3)+p(x_1&lt;x_2&lt;x_4)$ and  every $x$ are iid so both have same probabiltiy so is $2p(x_1&lt;x_2&lt;x_3)$ is right? But  i get the $p(x_1&lt;x_2&lt;x_3) = 1/6$ and then my answer is $1/3$ but the answer is $1/6.$ Can we write $P$ as $p(x_1&lt;x_2&lt;x_3&lt;x_4)+p(x_1&lt;x_2&lt;x_4&lt;x_3)$?
121345
My question comes from different modes of probability law, which I met. I see two options in equality: set belonged to $B(\mathbb R^n)$ and cartesian product of B belonged to $B(\mathbb R)$.
115716
Fix $\alpha , \beta &gt; 0$ with $\alpha + \beta = 1$ and consider the sequence of random variables defined by $X_0 = \theta \in (0,1)$ and  $$P(X_n = \alpha + \beta X_{n-1} \mid X_{n-1} , ... , X_0) = X_n , \qquad P(X_n =  \beta X_{n-1} \mid X_{n-1} , ... , X_0) =1- X_n.$$  I am trying to prove that $P(\lim_{n\rightarrow \infty} X_n = 1) = \theta$ and $P(\lim_{n\rightarrow \infty} X_n = 0) = 1-\theta$. It seems that there should be a way to do this using the result that $P(E \mid \mathcal F_n) \rightarrow P(E \mid \mathcal F)$, whenever $\mathcal F_n \uparrow \mathcal F$ is an increasing sequence of $\sigma$ algebras. But, I am not seeing a solution. Any suggestions?  By the way, this is an exercise on page 224 of Durrett's Probability: Theory and Examples.