5 votes
Accepted

Independence of diagnostic tests

$P(T_1 \cap T_2 | D)$ : Probability that the results are positive in both the tests given Fred has the disease. $P(T_1 \cap T_2)$ : Probability that the results are positive in both the tests taking ...
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4 votes
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X,Y, Z are independent then $X+Z$ and $Y+Z$ are independent

This is not true. Consider the following simple example: Experiment: Tossing a fair coin once. Sample Space: $S = \{H, T\}$ Define random variables $X$, $Y$ and $Z$ as $X(s)=0$ $\forall s \in S$ $Y(s)...
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  • 901
3 votes

What statement do I use here to get this equation $\Bbb{E}\left(S_k \cdot \Bbb{1}_{\{T=k\}}\right)=\Bbb{E}(S_k)\Bbb{E}(\Bbb{1}_{\{T=k\}})$?

The reason is that the $X_k$ and $T$ are independent, therefore the $S_k$ and the $\mathbf{1}_{\{T=k\}}$ are also independent, as both are functions of the $X_k$, in one side, and of $T$ in the other ...
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  • 26.6k
2 votes
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Probability of getting either Black or tabby

The answer is wrong. Let's visualize it: As you may see, there are 16 possibilities, and only highlighted 4 of them have no “red” in them (i.e., neither black, nor ...
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  • 2,830
2 votes

Probability of getting either Black or tabby

In your final part of the question, you should have added the two probabilities together, rather than multiplied them, the chances of getting either calico or ginger are $0.2+0.07=0.27$. However, it ...
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  • 6,082
2 votes

Probability of getting either Black or tabby

Also it seems like you were trying to calculate the probability of a black and a tabby kitten but the problem asks for the probability of a black or a tabby kitten. That could be two black, one black ...
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2 votes

X,Y, Z are independent then $X+Z$ and $Y+Z$ are independent

Take $X=0,Y=0$ and $Z$ any non-constant r.v. Then $X,Y,Z$ are independent but $Z=X+Z$ and $Z=Y+Z$ are not independent.
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  • 2,660
2 votes

Independence of two products of random variables with common factor

$X$ and $Y$ are not independent because $E(X^{2}Y^{2})=EA^{4}EB^{2}EC^{2}=EA^{4} \neq 1$ and $(EX^{2})(EY^{2})=EA^{2}EB^{2}EA^{2}EC^{2}=1$.
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  • 2,660
1 vote
Accepted

Exercise 1.9: Show that if a random variable X is independent of an event A, then a function of the random variable g(X) is also independent of A

$\int_A g(X(\omega))d\mathbb{P}(\omega) = \int_A\sum_{k=1}^n\alpha_k\mathbb{I}_{B_k}(x)d\mu_X(x) = \sum_{k=1}^n\alpha_k\int_A\mathbb{I}_{B_k}(x)d\mu_X(x)$ $ = \sum_{k=1}^n\alpha_k\mathbb{P}\{X\in \...
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  • 40.3k
1 vote
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Dependence of Events - Proof

If the events $A^{c}$ and $B^{c}$ are independent, then $A$ and $B$ are independent. Indeed, suppose that $A^{c}$ and $B^{c}$ are independent: \begin{align*} \mathbb{P}(A\cap B) & = 1 - \mathbb{P}(...
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1 vote
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What statement do I use here to get this equation $\Bbb{E}\left(S_k \cdot \Bbb{1}_{\{T=k\}}\right)=\Bbb{E}(S_k)\Bbb{E}(\Bbb{1}_{\{T=k\}})$?

You can see T as a stopping time. That is, you compute the average up to time T. The key idea, and in my opinion the easiest way to compute this, is to use the Law of Total Probability as follows, $$ ...
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1 vote
Accepted

Probability of union of independent Poisson point processes intersected with bounded set

Suppose $\{X_i\}_{i=1}^{\infty}$ are mutually independent Poisson Point Processes (PPPs) on the real number line with intensities $\{\lambda_i\}_{i=1}^{\infty}$. For any region $B \in \mathcal{B}(\...
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  • 20.3k
1 vote

If $A_1,A_2,…,A_n$ are independent then $A_1^c,A_2^c,…,A_n^c$ are also independent.

Partial attempt: Let $B_1 := \bigcap_{i=1}^k A_i^c$ and $B_2 := A_{k+1}^c$. \begin{align} P\left(\bigcap_{i=1}^{k+1} A_i^c\right) &= P(B_1 \cap B_2) \\ &= 1 - P(B_1^c \cup B_2^c) \\ &= 1 - ...
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  • 81.7k

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