Tag Info

Hot answers tagged

13

is it possible that the event has in fact zero probability? Yes. For a situation where this always happens, assume that one observes a random number $x$ drawn from the uniform distribution on $(0,1)$. Then the probability to observe $x$ is zero. (Proof: For every interval $I\subseteq(0,1)$ the probability to observe a number in $I$ is the length of ...


5

It is quite likely that you were told that $X$ and $Y$ are independent random variables, but neglected to pass on this information to us. Assuming that $X$ and $Y$ are independent Poisson random variables, $U = X+Y$ is also a Poisson random variable with parameter $\theta+\lambda$. Thus, the probability that $U = m$ ($m$ is a nonnegative integer here) is ...


5

Let $F(t)=e^{t/C}\int_t^\infty f(x)\,dx$. The stated assumption is equivalent to $F'(t)\le 0$ for large $t$. Therefore, $\int_t^\infty f(x)\,dx=O(e^{-t/C})$.


5

Yes. Since probabilities are nonnegative, we know that $P(A \cup B) \geq 0$ so that $-P(A \cup B) \leq 0$. Hence: $$ P(A \cap B) = P(A) + P(B) - P(A \cup B) \leq P(A) + P(B) + 0 = P(A) + P(B) $$ as desired.


5

$$ A\cap B\subset A\implies P(A\cap B)\leq P(A)=P(A)+0\leq P(A)+P(B). $$


5

All you need is the monotone convergence theorem. Note that if you write $\mathbb E(\cdot)$ as $\int (\cdot) d\mathbb P $ and also write $\sum_{i=1}^\infty(\cdot)$ as $\lim_{n\to\infty}\sum_{i=1}^n(\cdot)$, you can rewrite your desired statement as $$\int \lim_{n\to\infty} \sum_{i=1}^n X_n \ d\mathbb P = \lim_{n\to\infty} \int \sum_{i=1}^n X_n \ d\mathbb ...


5

Since you seem to be turning around this question and some of its variants again and again, let us try to answer it (almost) completely. First, as mentioned partially by the text you are reading, to know the characteristic function of every normal random vector, it is enough to know the characteristic function of a standard one-dimensional normal random ...


5

Starting where you stopped, assume that $E(X^+)$ is infinite, then $\sum\limits_{n=1}^\infty P(X_n\geqslant xn)=\sum\limits_{n=1}^\infty P(X\geqslant xn)$ diverges for every $x\gt0$ hence Borel-Canteli lemma implies that $X_n\geqslant xn$ for infinitely many $n$, almost surely, thus $\limsup\limits_{n\to\infty}\frac{X_n}n\geqslant x$, almost surely. This ...


4

If $x$ and $y$ are independent Gaussian random variables, then the linear combination $$z(t)= x\cos(wt)-y\sin(wt)$$ is also Gaussian. Can you calculate the mean and variance of $z(t)$?


4

In general, the inequality $$\mathbb{E}\|X-Y\|^2 \leq \mathbb{E}\|X-Z\|^2 + \mathbb{E}\|Z-Y\|^2$$ does not hold. Simply consider arbitrary $X$, $Z := \frac{1}{2} X$ and $Y = \frac{1}{4}X$. Then $$\mathbb{E}\|X-Y\|^2 = \frac{9}{16} \mathbb{E}\|X\|^2$$ whereas $$\mathbb{E}\|X-Z\|^2 + \mathbb{E}\|Z-Y\|^2 = \left( \frac{1}{4} + \frac{1}{16} \right) ...


4

First, $P(E_n)=1/n$ simply by symmetry, because $E_n$ happens when $\max\{X_1,X_2,\ldots,X_n\}$ is realized with $X_n$. Since the common distribution is continuous there is almost surely no ex aequo and $\max\{X_1,X_2,\ldots,X_n\}$ has as much chances to be realized with each $X_k$ with $1\leqslant k\leqslant n$, hence the result. Likewise, the ...


4

The mistake is that the number of possibilities for the second person (and subsequent ones) to pick a hat besides his depends on what's happened before. For example, if person number 1 picked hat number 2, there will be $n-1$ possibilities. Otherwise, $n-2$. I'm not sure how easy it is to give a proof along these lines that keeps track of these ...


4

If $X$ has finite first moment, then $$ \Bbb{E}(X) + \Bbb{E}(Y) = \Bbb{E}(X+Y) = \Bbb{E}(X), $$ so that subtraction of $\Bbb{E}(X)$ yields $\Bbb{E}(Y)=0$. Together with $Y\geq 0$ almost surely, this easily implies that $Y\equiv 0$ almost surely. EDIT: Without assuming that $X$ has finite first moment, we can use the following "truncation" trick: For $n ...


4

If in a box there are $n$ marbles, $k$ of them being red and $n-k$ being black, with $$\sum_{i}\binom{n-k}{i}\binom{k}{k-i}$$ you are counting the ways to select $i$ black marbles and $k-i$ red marbles, for any possible $i$. This is just the number of ways to select $k$ marbles out of $n$, i.e. $\binom{n}{k}$.


4

If $\mu$ has a finite fourth moment, use $E((XY)^2)=1=E(Y^2)$ and $E((XY)^4)=3=E(Y^4)$, hence $E(X^2)=E(X^4)=1$. This implies that $E((X^2-E(X^2))^2)=0$, thus $X^2=E(X^2)$ is almost surely constant. Since $E(X^2)=1$, $X=\pm1$ almost surely. Conversely, every Bernoulli distribution $\mu=p\delta_1+(1-p)\delta_{-1}$ is a solution. Otherwise, use $E(\mathrm ...


4

What you are asked to show: If $\mathbb P(A_n)\lt1$ for every $n$ and $\prod\limits_{n=1}^\infty \mathbb P( A_n^c)=0$ then $\sum\limits_n\mathbb P( A_n)$ diverges. Thus, Borel-Cantelli lemma is not involved in the proof that the series $\sum\limits_n\mathbb P( A_n)$ diverges, which is purely a problem of real analysis. In full generality: Consider ...


4

The idea is that, since the series $\sum\limits_nP(X_n\ne-1)$ converges, Borel-Cantelli lemma (the simple one) indicates that, almost surely, there exists some finite $N$ such that $X_n=-1$ for every $n\geqslant N$. Every such sequence $(X_n)$ is such that $S_n/n\to-1$ hence $P(S_n/n\to-1)=1$. The strong law of large numbers does not apply, although $(X_n)$ ...


3

Your question is not a mathematical question, since it depends on the interpretation of probability. It also depends on precisely what you mean by "given that we have observed...". For example, subjective Bayesians think of probability theory as theory of how (idealized) rational agents should update their beliefs in light of new observations. It is part of ...


3

Assume that $Y:(\Omega,\mathcal F)\to(E,\mathcal E)$, that is, that the random variable $Y$ is $E$-valued (in most of the cases, $(E,\mathcal E)=(\mathbb R,\mathcal B(\mathbb R))$ but the natural setting of the question is more general). On the other hand, the random variable $X$ must be real valued, thus, $X:(\Omega,\mathcal F)\to(\mathbb R,\mathcal ...


3

The high road to this result, which explains at the same time your second question, is to note that $$E[(X-E[X])^2]=\inf\{E[(X-a)^2]\,\mid\,a\in\mathbb R\},$$ while $$E[(X-E[X|G])^2]=\inf\{E[(X-Y)^2]\,\mid\,Y\in L^2(G)\},$$ hence the latter, being an infimum on at least as many random variables, is not larger than the former. Edit: Pushing a little the idea ...


3

The transition rate matrix is another name for the generator matrix for a continuous time markov chain. A transition matrix is for a discrete time markov chain. $M$ would be for a continuous time markov chain (the underlying process for the queue is based on Poisson processes) while $T$ would be for a discrete time markov chain (the underlying process ...


3

More generally... one may want to keep in mind that Markov chains (in discrete time) and Markov processes (in continuous time) are different (although related) objects. 1. A Markov chain $(X_n)$, indexed by $n$ integer, is described by a transition matrix $P$, such that, for every states $(x,y)$ and every $n$, $$\Pr(X_{n+1}=y\mid X_n=x)=P_{xy}.$$ Thus: ...


3

Tightness (of a family of measures, not of a convergence) is useful to deduce the convergence in distribution (when $x\to\infty$, say) of a whole process $X_x=(X_x(t))_t$ from the convergence in distribution of all its finite-dimensional marginals. Thus, the somewhat sloppy "In order to show that $X(t) \xrightarrow{d} Y(t)$, we need to show tightness ...


3

No, this is not true. First of all, in general $E[X \mid \mathscr{G}]$ is only defined up to sets of measure zero, so $E[X \mid \mathscr{G}](A)$ is not a well defined set. But setting this aside, try a probability space with two points: $\Omega = \{a,b\}$ with $\mathscr{F} = 2^\Omega$ and $P(\{a\}) = P(\{b\}) = 1/2$. Set $X(a) = 1$ and $X(b) = -1$ so that ...


3

First part: For every $n$, the simple Markov property at time $nK$ yields $$P_j(T_i\gt(n+1)K\mid T_i\gt nK,X_{nK}=k)=u(k),\qquad u(k)=P_k(T_i\gt K),$$ and $u(\ )\leqslant1-\varepsilon$ uniformly by hypothesis hence $$P_j(T_i\gt(n+1)K,X_{nK}=k)\leqslant(1-\varepsilon)P_j(T_i\gt nK,X_{nK}=k).$$ Summing these over $k$ yields ...


3

CLT does not apply to the present setting because (the simplest version of) CLT assumes that $(X_n)$ is i.i.d. while here the distribution of $X_n$ depends on $n$. With Borel-Cantelli I obtained; $$\sum_n\Pr(X_n\neq0)<\infty\Longrightarrow\Pr(\limsup\limits_n\{X_n\neq0\})=0$$ What does it mean now? The set $\limsup\limits_n\{X_n=0\}$ has probability ...


3

Considering that $\phi(0)=1$ and that $\phi_k(0)=1$ for every $k$, one sees that the condition $$\sum_kp_k=1$$ is necessary. To prove that is is also sufficient and to avoid most of the technicalities, consider some random variables $N$ and $(X_k)$ defined on the same probability space, $N$ independent of $(X_k)$, each $X_k$ with characteristic function ...


3

It is not totally obvious to me what you are asking for, but I would have thought the argument is something like $$\Pr(R \le \mu + \sigma \Phi^{-1}(u)) = \Pr(\epsilon \le \Phi^{-1}(u)) = \Phi(\Phi^{-1}(u)) =u$$


3

Hint: Apply Itô's formula to $$Z_t := \text{arsinh} \, X_t = \int_0^{X_t} \frac{1}{\sqrt{1+y^2}} \, dy.$$ General approach: The given SDE is an autonomous, i.e. the coefficients do not depend explicitly on the time: $$dX_t = b(X_t) \, dt + \sigma(X_t) \, dW_t. \tag{1}$$ In some cases, these SDEs can be transformed into linear SDEs - and since linear SDEs ...


3

You only need $X$ is $\mathcal{F}_2$-measurable, which gives: $$ E[XY|\mathcal{F_2}]=X E[Y|\mathcal{F}_2]. $$ Now, using $\mathcal{F}_1\subset\mathcal{F}_2$ and iterated conditioning for the first equality below, we have $$ E[XY|\mathcal{F}_1]=E[E[XY|\mathcal{F_2}]|\mathcal{F}_1]=E[XE[Y|\mathcal{F_2}]|\mathcal{F}_1]. $$



Only top voted, non community-wiki answers of a minimum length are eligible