# Tag Info

## Hot answers tagged average

14

Summary: The expected number of survivors after a shootout given as $$\lim_{n\rightarrow\infty}\operatorname{E}[n]\approx 0.284051\ n;\quad \text{(Tao/Wu - see below)}$$ is, if not correct, almost certainly very close to being correct (see update 2). However, this is disputed by Finch in Mathematical constants (again, see below for details). The results ...

4

It's called the weighted average. :) Seriously: this is exactly how generalized weighted averages are defined. Nice work inventing it for yourself!

3

The average velocity is simply the distance, covered by the object, divided by the time it took for the object to cover that distance. There is no need to calculate $f'(t)$ to answer (a,b,c). You only need to calculate $\frac{f(t_1-f(t_0)}{t_1-t_0}$ where $t_0=1$ and $t_1$ takes three different values.

2

The centroid of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is the point whose cartesian components are the means of the individual $x$-coordinates and $y$-coordinates: $$(x_C, y_C) = \frac{1}{3} (x_1 + x_2 + x_3, y_1 + y_2 + y_3)$$ The centroid of a triangle can also be easily constructed by bisecting each side of the triangle and then ...

2

Consider a random variable $X_i$ such that $X_i = 1$ if the $i^{th}$ ball is red after 100 minutes. Note that $Pr(X_i = 1) = 1 - ({99 \over 100})^{100}$. Using linearity of expectation, $$E(X) = \sum_{i=1}^{100} E(X_i) = 100\left(1 - \left({99 \over 100}\right)^{100}\right) \approx 63.4$$

2

Fix $b$, then the average value of $a^b$ is $\int_0^1 a^b da=1/(b+1)$. Then the average value of $1/(b+1)$ is $\ln2$. On the other hand, as Jano says in comments, I don't know whether the order of integration matters.

2

The expected value is $\int_Q a^b dA$, where $Q$ is the unit square $[0,1]\times[0,1]$ (define $0^0=1$ if you must). Since everything is well-behaved, $\int_Q a^b dA=\int_0^1\int_0^1 a^b \,db \,da=\int_0^1\int_0^1 a^b \,da\,db$. You can now proceed as in Michael's answer.

2

Instead of considering $100$ cards, we assume there are $n$ cards, where $n \geq 1$. Let $X_n$ denotes the # of cycles given $n$ cards and we are interested in the following identity $\mathbb{E}[X_n]$. We also introduce another random variable, denoted as $L$, which denotes the length of the first cycle, thus $L \in [1, n]$. We use the following fact of ...

2

Density of a sum is convolution of densities $$f_{X_1+X_2}(z)=\int f_{X_1}(y)f_{X_2}(z-y)dy$$ To see why support of $Y$ doesn't have to be $G$, consider $X_i=Binomial(1,1/2)+Uniform(-\epsilon,\epsilon)$ for $\epsilon<1/3$. Then $Y$ has non-zero density on the neighborhood of $1/2$ for even $n$ while density of $X$ is zero on that neighborhood. In general, ...

1

The problem is that the first elements of the sequence may not be close to $a$. So you have to deal with the elements that are not close enough to $a$. Recall that the definition of $\lim_{n\to\infty} a_n = a$ is that $$\forall \epsilon > 0, \exists N \mbox{ such that } \forall n \geqslant N, \,\, |a_n - a| < \epsilon.$$ So for $n < N$, the ...

1

HINT: Write $$\frac1n \sum_{k=0}^na_k=\frac1n \sum_{k=0}^Na_k+\frac1n \sum_{k=N+1}^na_k \tag 1$$ For and $\epsilon>0$, choose $N$ so that $a-\epsilon <a_k<a+\epsilon$ for $k>N$. Then for this fixed $N$, observe that the first sum on the right-hand side of $(1)$ can be made arbitrarily small by choosing $n$ large enough. For the second sum, ...

1

Here's a simulation in Octave. "Model" is the linear model $\frac{2}{7}n$ which martin showed above. Maybe it would be interesting to also investigate the spread and how it changes with $n$? Even if the mean value of surviving gangstas is pretty close to $\frac{2}{7}n$ as martin discovered in his answer, we can also see that there is a quite large spread ...

1

As has been noted in comments, the expected number of attempts to get one item that has probability $p$ per attempt is $1/p$. To find the expected number of attempts for a collection of items, we can use inclusion-exclusion. Let $A_n$ denote the event that the loot from $n$ kills doesn't contain $3$ pants, $B_n$ the event that it doesn't contain $2$ ...

1

This is indeed the case in general: if we take the average of each subset of $\{1,2,\ldots,n\}$ and then average those averages, we find the average $\frac{n+1}{2}$ of the numbers $1$, $2$, ..., $n$. In fact, something stronger is true: the average already equals $\frac{n+1}{2}$ if we only consider the $k$-element subsets of $\{1,2,\ldots,n\}$ for some $1 ... 1 Both answers of lhf and Michael are, of course, correct. Though, to complete them, I looked into the limit case for which you ask in your question, that is for a sequence of independent$U_1, U_2, \dots, U_n$uniformly distributed over$(0,1)$, what is the expectation of$(((U_1^{U_2})^{U_3})\dots)^{U_n} = U_1^{{U_2}{U_3}\dots{U_n}}$? We know that for each ... 1 It'is always the case as: $${1\over n} \sum \limits_{i=1}^n {1 \over m} \sum \limits_{j=1}^m (a_i - b_j)={1\over n} \sum \limits_{i=1}^n {1 \over m} (\sum \limits_{j=1}^m a_i - \sum \limits_{j=1}^m b_j)={1\over n} \sum \limits_{i=1}^n {1 \over m} (m a_i - \sum \limits_{j=1}^m b_j)={1\over n} \sum \limits_{i=1}^n (a_i - {1 \over m}\sum \limits_{j=1}^m b_j)={1 ... 1 Notice, we can directly apply the formula for average velocity (\bar{v}). the average velocity of the object is given as$$\bar{v}=\frac{\text{total distance traveled in time interval}\ \Delta t}{\text{total time required}\ (\Delta t)}=\frac{f(t_1)-f(t_2)}{t_2-t_1}=\frac{f(t_1)-f(t_2)}{\Delta}$\$ Since, the height of the object from the ground is ...

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