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What can I say about $P(B|A_1 \cap A_2)$ if I know that $P(B|A_1), P(B|A_2) \approx 1$?

The event $A_1\cap A_2$ may be much smaller than either of the events $A_1$ or $A_2$. So even if most outcomes of the events are also outcomes of event $B$, that does not require that any outcomes of ...
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Computing conditional expectation from a given random variable

\begin{gather*} P_{X/\mathcal{G}}(A)=\frac{P((X\in A)\cap\{a\})}{P(\{a\})}I_{\{a\}}(X)+\frac{P((X\in A)\cap\{b,c\})}{P(\{b,c\})}I_{\{b,c\}}(X)\\ E[X/\mathcal{G}](\omega)=\sum_{i=1}^3iP_{X/\mathcal{G}}(...
Speltzu's user avatar
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1 vote

What can I say about $P(B|A_1 \cap A_2)$ if I know that $P(B|A_1), P(B|A_2) \approx 1$?

The general principle that you are suggesting is not true. Here is an example. Suppose I am the greatest chess player that ever lived. I have two students. They will be attending a conference with 998 ...
Jason Swanson's user avatar
1 vote
Accepted

Computing conditional expectation from a given random variable

The conditional expectation must be measurable with respect to $\mathcal F,$ therefore the inverse image of $Y(b)$ must belong to $\mathcal F = \{\emptyset,\{a\},\{b,c\},\Omega\}.$ Since the inverse ...
Lieven's user avatar
  • 850
1 vote
Accepted

Conditional Independence involving four events

When $A, B, C, H$ are events, this property seems wrong. For example, when $H = \Omega$, then it amounts to say "if $A$ and $C$ are independent, $B$ and $C$ are independent, then $A \cap B$ and $...
Zhanxiong's user avatar
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1 vote

Conditional probability density function of the parameter $\lambda$, given a data set.

Yes, you are updating a prior distribution of $\lambda$ to a posterior distribution given the data and $f(\lambda)$ is the prior. There is a problem with your denominator though. The $X_i$ are not ...
spaceisdarkgreen's user avatar
0 votes

Prove $\mathbb{P}(\tau > kN) \leq (1-\varepsilon)^k$ if $\mathbb{P}( \tau \leq n+N|\mathcal{F}_n) > \varepsilon$

I recommend thinking about what each of these probabilities means intuitively. The first is "probability we're still going after $kN$ steps". The second is "given that we're still ...
Ziv's user avatar
  • 131
0 votes

Limit of expectation of product of random variables - using conditional expectation

The statement is false. Let $X$ have uniform distrbution on $(-1,1)$, $X_n=X=Z$ for all $n$. Then $EX_nZ=\frac 1 3$ for all $n$.
geetha290krm's user avatar
  • 38.2k
1 vote

Convergence of bootstrap variance estimator

Let $(\Omega,\mathcal F, P)$ be the underlying probability space, $X_1, X_2, ..., X_n$ iid copies of an real-valued random variable $X$ with non-negative finite second moment $\sigma^2$. I denote the ...
Syd Amerikaner's user avatar
1 vote
Accepted

Determine the distribution of $X$

The law of total probability (LoTP) helps you compute the probability of an event by summing over simpler pieces, where the conditional probability of each piece is either given or is easy to compute. ...
Michael's user avatar
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0 votes

Resampling techniques to obtain uniform random variables.

Your triangular distribution has density $f_X(x)=\dfrac{h-|x|}{h^2}\mathbb I_{[|x|\le h]}$. There are various approaches. The simplest is: If $h>1$, you could resample by accepting each $x_i$ with ...
Henry's user avatar
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0 votes

Probability of drawing a red ball from a bag given bag state after ball is drawn

Let's look at how we can form the initial configuration: For any ball in the bag, there is a constant probability of p to pick one red ball and q to pick a blue ball; and we know p+q = 1. When we know ...
Herbert The Bird's user avatar
1 vote

Why is my approach wrong for this probability question?

\begin{align} & \color{#44f}{\sum_{m = 1}^{40}{1 \over 40}\sum_{n = m + 1}^{60}{1 \over 60}} = {1 \over 2400}\sum_{m = 1}^{40}\left(60 - m\right) = {60 \times 40 - 40 \times 41/2 \over 2400} \\[...
Felix Marin's user avatar
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3 votes

Why is my approach wrong for this probability question?

I completely agree with your initial analysis, and (in fact) the case of $~p(X_{60}) > 40~$ is completely dealt with by assigning this the probability of $~(1/3).$ Your mistake was in overlooking ...
user2661923's user avatar
  • 36.1k
1 vote

Can these problems be solved using the same technique?

A point of clarification: For problem $2$, I am reading the problem as ignoring all but $13$ cards (the four Kings, Queens, Jacks, and the $2\heartsuit$). Thus, I'd say that the path $9\spadesuit, 3\...
lulu's user avatar
  • 71.1k
0 votes

Can these problems be solved using the same technique?

For the second problem $P(\operatorname{desired})=\frac{4}{52}×\frac{4}{51}×\frac{4}{50}×6×\frac{1}{49}$
MathStackexchangeIsNotSoBad's user avatar
2 votes

Probability of $r$ consecutive heads occurring exactly once in infinite tosses

To answer the first question being asked in the post: no, the reformulation is not equivalent—in part because the original question asks for a probability, while the reformulation sets up a game but ...
Greg Martin's user avatar
1 vote

Geometric Brownian Motion problem - Compute $\mathbb{P}(X_3 < 3)$

Whereas $X_{t} \sim GBM(\mu, \sigma^2)$, another way to solve it is to remember that $$ X_{t+s} = X_{t} \ e^{\left( \mu-\frac{\sigma^2}{2} \right)s \ + \ \sigma(W_{t+s} - W_{t})}, $$ then $$X_{1+2} = ...
Victor Nunes's user avatar
0 votes

Probability of Drawing a Red Ball - Revisited - $~p(A|B) = \frac{p(A,B)}{p(B)}.$

For $~k \in \{1,2,\cdots,5\},~$ let $~f(k)~$ denote the probability of drawing 2 red balls, one at a time, with replacement, under the assumption that there are $~k~$ red balls, out of the $~5~$ balls....
user2661923's user avatar
  • 36.1k
1 vote

Intuition for Symmetry while calculating Mean of Hypergeometric Distribution

Imagine that all the $N\; (=b+r)$ marbles are sequentially numbered $1\;thru\; N$. Now all sequences of numbers are equally likely, so the probability that a paticular numbered ball is at position i ...
true blue anil's user avatar
6 votes
Accepted

Show that if $E(X_n|\mathcal{F}_n)\rightarrow_p 0$ then $X_n\rightarrow_p 0$.

Here is a hint: If $X$ is non-negative and $\mathcal{F}$ is a sub-$\sigma$-field, then by the concavity of the function $x\mapsto\min\{1,x\}$ and the conditional Jensen's inequality, $$\mathbb{E}[\min\...
Sangchul Lee's user avatar
0 votes

Ordered Chain Rule of Probability

The issue in the tree is that when you write $P(A,B)$ what you should be writing is $P(A_1,B_2)$ - if the sampling is done without replacement, $P(A)$ can be different at each set of branches...
Red Five's user avatar
  • 2,593
4 votes
Accepted

How to determine which event is the conditional event?

It's neither. For conditional probability you assume that one event happened, and then ask for the probability that an additional event also happened. Here, you aren't assuming either event. Here ...
lulu's user avatar
  • 71.1k
1 vote
Accepted

Question 29 from Chapter 3 of A first course to probability from Sheldon Ross ed. 10

What you have calculated for first strategy and your assumption are both correct but there is an error in your answer for strategy $2$. In $2$nd strategy for a $3$ occupant house member to be chosen ...
Yash Shrivastava's user avatar
1 vote
Accepted

Question 11 from chapter 3 of Sheldon Ross book A First course to probability 10th ed.

The mistake is in your calculation of $P(A_s)$. You forgot that the ace of spades can be picked on either turn, so $$P(A_s)=\frac1{52}+\frac {51}{52}\cdot\frac1{51}=\frac1{26}.$$
Ted Shifrin's user avatar
1 vote
Accepted

Density Functions and Probability Calculation

I don't need help solving this integral, I just wanna know how to get here. $$\int_{0.5}^1\int_{1.5-y}^11\,\mathrm d x\,\mathrm d y$$ The support is $\{(x,y):0\leq x\leq 1~,~ 0\leq y\leq 1\}$ and the ...
Graham Kemp's user avatar
1 vote

Conditional Probability in Selection of Nuts from Different Weight Distributions

I will just find the probability of picking a nut from Bavaria (given its weight is >10) since you can then do the others. You must use Bayes theorem $$ P(B|X>10) = \frac{P(X>10|B)P(B)}{P(X&...
Daniel Adams's user avatar
2 votes
Accepted

Conditional expectations preserve convergence in measure

Easy counter-example: Suppose $X$ and $X_n$ are independent of $\mathcal G$. Then you are asking if $X_n \to X$ a.s. implies $EX_n \to EX$. A standard counter-example is $X_n=n1_{(0,\frac1 n)}, X=0$ ...
geetha290krm's user avatar
  • 38.2k
1 vote
Accepted

Once $X\sim \text{gamma}(\alpha = 12, \beta = 2)$ is observed, $Y$ is randomly chosen from $(0,x)$. Evaluate $E(Y)$

From 'randomly chosen from $(0,x)$' we infer that, given any fixed value $x$ that $X$ can take, $Y$ follows a uniform distribution in $(0,x)$. This means that $E[Y\,|\,X=x]=\dfrac{x}{2}$, and so $E[Y\,...
Julio Puerta's user avatar
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