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3

The full equation reads: $$P(\text{rain}) = P(T>25)\cdot P(\text{rain if } T>25) + P(T\le 25)\cdot P(\text{rain if } T\le 25).\tag 1$$ The problem statement explicitly contains: The temperature at noon is equally likely to be above 25°C, or at/below 25°C. That is, $P(T>25) = P(T\le 25)$. Since $P(T>25) + P(T\le 25) = 1$, it immediately ...


3

Hint: use the Borel-Cantelli lemma to show that $$P(X_n \ne 2 \text{ i.o.}) = 0.$$ (Independence is not needed,)


2

To complete the proof you need to show that $\lambda_n\rightarrow \lambda$ implies $(1+\lambda_n/n)^n\rightarrow e^{\lambda}$. Here's the proof: $$(1+\lambda_n/n)^n=\exp(n\log(1+\lambda_n/n)).$$ Now expand the log: $$n\log(1+\lambda_n/n)=n\left(\frac{\lambda_n}{n}+O(\lambda^2_n/n^2)\right)=\lambda_n+nO(\lambda^2_n/n^2).$$ Notice that $n\cdot ...


2

As suggested by @Dominik, we can adopt the usual proof to get the inequality even for non-centered random variables. The point is that the expectation "magically" cancels out. We assume (as before) that $$ \mathbb{E}\left|X_{\ell}\right|^{n}\leq\frac{n!}{2}\cdot R^{n-2}\cdot\sigma_{\ell}^{2} $$ for all $1\leq\ell\leq m$ and $n\geq2$. In one direction, we ...


2

Without brute force: $$ \begin{aligned} E[X_1X_2|X_2X_3] &= E[E[X_1X_2|X_2X_3,X_2]|X_2X_3] \\ &= E[X_2E[X_1|X_2X_3,X_2]|X_2X_3] \\ &= E[X_2E[X_1]|X_2X_3]\\ &= E[X_2\cdot 0 |X_2X_3]\\ &=0 \end{aligned} $$ First equation is the tower property for conditional expectations, ...


2

The random variable is not limited to values less than $x$. For instance, I can show you that the function $$ X(t) = \frac{1}{t} $$ is a measurable function on $(0, 1)$. Here's how. Let's look at $$ \{ t \mid X(t) \le 11 \} $$ That's the set of all points in the domain for which $X(t) = 1/t$ is less than 11, which is exactly $$ A = \{t \mid \frac{1}{11} ...


2

A random variable $X$ on $\Omega$ is no more and no less than a function $X:\>\Omega\to{\mathbb R}$ satisfying the technical condition that it is measurable: For any $x\in{\mathbb R}$ the set $\{\omega\in\Omega\>|\>X(\omega)\leq x\}$ belongs to ${\cal F}$. This guarantees that for any two given values $a$, $b$ the probability $$P[a\leq X(\omega)\leq ...


2

Let $Y_i = |X_i|$, and let $s_i$ be chosen randomly for $\{\pm 1\}$. The variables $Y_i$ are iid and have a certain distribution. One way of generating the $X_i$ is generating the $Y_i$ and $s_i$, and using the formula $X_i = s_i Y_i$. We will show that for every $y_1,\ldots,y_n$, $$ P(|S_n| \geq \max_i y_i \; | \; \vec{Y} = \vec{y}) \geq 1/2, $$ which ...


2

You didn't apply the integration by parts formula correctly, e.g. $$\int f'(x)b(x)p(x,t|y)dx=uv-\int v\,du=f(x)b(x)p(x,t|y)-\int f(x)\partial_x\left[b(x)p(x,t|y)\right]dx$$ is not correct (note that the right-hand side depends on $f(x) b(x) p(x,t \mid y)$ whereas the left-hand side does not depend at all on $x$). Integration by parts gives ...


2

The English in the problem is not as clear as it might be. I think (2) asks for the waiting time to get HH (two Heads in a row). From the referenced page on this site you have found that the answer is 6. Or, equivalently (2) could be interpreted as the waiting time to get TT. By symmetry (fair coin) this as also 6. I think (1) asks for the waiting to get ...


2

Consider a "spinner": an object like an unmagnetized compass needle that can pivots freely around an axis, and is stable pointing in any direction. You give it a spin and see where it comes to rest, measuring the resulting angle (divided by $2\pi$) as a number from $0$ to $1$.


2

Although it cannot be seen clearly, I think the steps you have problem with is this, $$\int{(f+g)}d\mu=\inf_{\psi\leq f+g}\int\psi d\mu$$ $$\leq\inf_{\psi_1\leq f, \psi_2\leq g}\left(\int\psi_1 d\mu+\int\psi_2 d\mu\right).$$ Just note that $\psi_1\leq f$ and $\psi_2\leq g \implies\psi=\psi_1+\psi_2\leq f+g$. Set-theoretically, we have the following ...


1

When the sequence of event $(A_n)_n$ is increasing or decreasing this propriety is true.


1

The field of probability can be made mathematically rigorous. Introductory textbooks on probability tend to use terminology like 'sample space', 'outcome', 'event', and outcomes are given names like 'Heads' or 'HTT' or 'King of Spades', all in an attempt to keep things informal and intuitive. These texts often refrain from defining such concepts precisely, ...


1

The number of sequences in which the minimal integer $i$ appears exactly $e$ times is $$ \binom ne (k-i)^{n-e}, $$ since there are $\binom ne$ ways to decide where the $i$s go and $k-i$ choices of larger integers for each other spot. The quantity you want is then the sum of this over all $e\ge j$. Equivalently, $$ p_{ij} = \frac1{k^n} \sum_{j\le e\le n} ...


1

due to symmetry, half the women are shorter that 165, so you need to calculate $$p(M<165)=p\left(z<\frac{165-178}{8}\right)$$ Can you finish it?


1

I would break this question up into two pieces. First, what height are half of the women taller than? Since the mean of the women distribution is $165$ and the median of a normal distribution is its mean, the answer is 165. The second half of the question is: What proportion of men are less than the height we just calculated. i.e. What proportion of men ...


1

You need the cumulative distribution function $P(X\leq k)$ in order to approximately represent the quantity you are interested in, i.e., winning in up to $k+1$ tickets. I say approximately since the geometric models sampling without replacement, but for this case of large n (say ${49 \choose 6}$) and tiny $p=1/n$ it doesn't matter.


1

In analog-to-digital conversion a quantization error occurs. This error is either due to rounding or truncation. When the original signal is much larger than one least significant bit (LSB), the quantization error is not significantly correlated with the signal, and has an approximately uniform distribution. The RMS error therefore follows from the variance ...


1

Let's redefine $x^*$ according to my comment above: $$ x^* = \inf\left\{ y : \int_0^T 1_{\{x(t)\leq y\}} dt \geq \tau\right\} $$ Assume $x(t)$ takes values in a state space of nonnegative integers, so $x(t) \in \{0, 1, 2, \ldots\}$ for all $t$. Assume that $x(0)=0$. Then the infimum value is achieved by a particular integer $x^* \in \{0, 1, 2, \ldots\}$ ...


1

I think you misspoke somewhat in your first statement: first hitting times are stopping times, but there are many other kinds of stopping times. What you've written is a last exit time, which is not a stopping time, because as you've said we cannot know whether $\tau \leq t$ by observing the process up to time $t$. This creates some difficulties in ...


1

Let $R$,$G_{25}$,$L_{25}$, denote the events that it rains in the afternoon, the temperature is greater than $25$, the temperature is less than $25$, respectively. We have the following data: $P(R)=0.6$, $P(G_{25})=P(L_{25})=0.5$, $P(R\mid L_{25})=0.4$. The question is $$P(R\mid G_{25})=\frac{P(R\cap G_{25})}{0.5}=2(P(R)-P(R\cap L_{25}))=$$ ...


1

This says The occurrence or non-occurrence of every event whose occurrence or non-occurrence is determined by the values of $X_1,\ldots,X_n$ is determined by the values of $Y_1,\ldots,Y_n$; and The occurrence or non-occurrence of every event whose occurrence or non-occurrence is determined by the values of $Y_1,\ldots,Y_n$ is determined by the values of ...


1

The player meets monsters at the rate $\lambda$, and with probability $p$ they are winning monsters, so the player meets winning monsters at the rate $\lambda p$. We can extend the game up to $t=1$, with monsters appearing after the player's death meeting a dead player. Then, since the Poisson monsters appear uniformly in time, the probability distribution ...


1

I think for any realization of random variable : $ X_{1},...,X_{n}$, the empirical distribution $F_{n}(x)=\frac{1}{n}\sum_{i=1}^{n}I_{\{X_{i}\le x\}}$ is just a discrete distribution function concentrated on these n value and attached weight $\frac{1}{n}$ to each of them. Thus, the integral $\int g(x)dF_{n}(x)$ is same as expected value of $g(x)$ with ...


1

Perhaps the quickest way to see that $A$ and $B$ aren't in general conditionally independent given $C\cap D$ is to take the entire space for $C$. Then $A$ and $B$ being conditionally independent given $C$ is equivalent to $A$ and $B$ being independent, and $C\cap D=D$. Certainly not all independent events are conditionally independent given arbitrary $D$.


1

The actual definition is that a function $X:\Omega\to\mathbb R$ is called a random variable when $X$ is $(\mathcal F, \mathcal B(\mathbb R))$-measurable. In other words, $X^{-1}(B)\in\mathcal F$ for any $B\in\mathcal B(\mathbb R)$, where $\mathcal B(\mathbb R)$ is the Borel $\sigma$-algebra on $\mathbb R$. Now, it is sufficient that ...


1

Let's have three biased coins. Coin $A$ produces heads with probability $\alpha$, coin $B$ produces heads with probability $p$ and coin $C$ produces heads with probability $q$. Here is the experiment: We flip coin $A$. If the result is heads then we flip coin $B$ otherwise we flip coin $C$. The result of the experiment is the result of the second flip. We ...


1

"A is independent of C", $A\perp C$, means that: $\mathsf P(A\cap C) = \mathsf P(A)\;\mathsf P(C)$ Likewise, the meaning of "A conditioned on B is independent of C," is that "A is conditionally independent of C, when given B". $$A\perp C \mid B \;\iff\; \mathsf P(A\cap C\mid B) = \mathsf P(A\mid B)\;\mathsf P(C\mid B)$$ Given the definition above, ...


1

Consider $$\tau_n = \frac{[2^n\tau] + 1}{2^n}$$ Note that $\tau_n$ is discrete and $\tau_n \downarrow \tau$. Now use the result you have for the discrete cases $$P(X_{\tau_n}>c\vert \tau_n=j)=\dfrac{P(X_{\tau_n}>c, \tau_n=j)}{P(\tau_n=j)}=\dfrac{P(X_{j}>c, \tau_n=j)}{P(\tau_n=j)}=P(X_{j}>c). \quad j \in \Bbb{Z}_+/ 2^n$$ Now to the continous ...



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