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## algorithms – A successful search takes \$Theta(1 + alpha)\$ time on average when resolving collisions by chaining

I would to discuss a proof found in CLRS book please.

Theorem: In a hash table in which collisions are resolved by chaining, a successful search takes time $$Theta(1 + alpha)$$, on the average, under the assumption of simple uniform hashing.

Proof: Assume we $$n$$ elements are equally likely to be searched in the table. The number of elements examined during a successful search for an element $$x$$ is 1 more than the number of elements that appear before $$x$$ in $$x$$’s list. New element will be added at the end of the list.

To find the expected number of elements examined, we take the average, over the n elements $$x$$ in the table, of 1 plus the expected number of elements added to $$x$$’s list after x was added to the list. Let $$x_i$$ denote the ith element inserted into the table, for $$i = 1, 2, cdots, n$$ and let $$k_i = key(x_i)$$. For keys $$k_i$$ and $$k_j$$, we define the indicator random variable $$X_{ij} = I {h(k_i) = h(k_j)}$$. Under the assumption of simple uniform hashing, we have $$Pr{h(k_i) = h(k_j)} = 1/m$$, $$E (X_{ij}) = 1/m$$. Thus, the expected number of elements examined in a successful search is

$$begin{gather} Eleft( frac{1}{n} sum_{i=1}^{n} left( 1 + sum_{i=1}^{n} X_{ij} right)right) tag{1} label{1}\ =frac{1}{n} times left(sum_{i=1}^{n} left( 1 + sum_{i=1}^{n} Eleft(X_{ij}right) right)right) &text{(Based on Linearlity of Expectation)}\ vdots \ =Theta(1+alpha) end{gather}$$

Problems:

1. The number of elements examined during a successful search for an element $$x$$ is 1 more than the number of elements that appear before $$x$$ in $$x$$’s list. So, if our table is empty and we have only 1 element, then the time needed would be $$O(1)$$. If all elements are hashed to first entry of the table, then we would need $$O(Elements)$$ to find the item as I understand. Why the time needed is $$Theta(1+alpha)$$ please?
2. How equation ref{1} was brought in first place? From what I understood, since all elements to be found are equally likely, then we would have $$frac{1}{n}$$, so probably this is why its their in ref{1}.

## worksheet function – How to find average with my condition in Excel?

Let’s say that this is my bank statement:

Sl. No. Tran Date Withdrawal Deposit Balance Amount
1 01-07-2020 100 100
2 02-07-2020 5 95
3 03-07-2020 500 595
4 06-07-2020 50 545
5 06-07-2020 8000 8545
6 06-07-2020 3000 5545
7 31-07-2020 5 5540
8 31-07-2020 10000 15540

Now I want to find out my monthly average balance.

To find it, I must add all the end of day balances and then divide by the number of days. But, the bank statement does not have end of day balances and just lists the transactions and their respective dates. Days where no transaction took place is not mentioned in the statement.

Now, let’s say that I have an Excel sheet of my bank statement. How can I find the monthly average balance? I read something about AVERAGEIF but I am not sure if it will be helpful in my case.

## pr.probability – Average value of \$frac{x’A^2x}{x’A^3x}\$ over surface of \$n\$-dimensional sphere

Suppose $$A$$ is a diagonal matrix with eigenvalues $$1,frac{1}{2},frac{1}{3},ldots,frac{1}{n}$$ and $$x$$ is drawn from standard Gaussian in $$n$$ dimensions. Define $$z_n$$ as follows
$$z_n=E_{xsim mathcal{N}left(0, I_nright)}left(frac{x^T A^2 x}{x^T A^3 x}right)$$

Is it possible to prove or disprove the following?

$$lim_{nto infty} z_n = 2$$

This is a crosspost from math.SE where several people provided altnernative characterizations of $$z_n$$ but which don’t quite settle the question.

Motivation: $$z_n$$ is the expected value of learning rate which maximizes loss decrease for a gradient descent step on a quadratic $$A$$ and random starting point. If the limit is 2, this would be a nice mathematical illustration behind the heuristic used in practice, “in high dimensions — set learning rate as high as possible”

## database design – In dolphindb, how to take the average of the top 5% records in a table

For example, sort by the trademoney,then take the average, minimum and standard deviation of the top 5%. like each stock is calculated by group, now i got SecurityID and TradeMoney of each stock,and 5day level 2 data, which is about 30 million.

how to take the average of top 5% records in this table.

## How can I derive an expression for average case key comparisons in merge sort?

I know that the worst case is 𝑛log𝑛−(𝑛−1) and the best case is 𝑛log𝑛/2. Is there an expression for the average case?

## dnd 5e – What is the average damage of a Potion of Poison?

Let us assume that our victim has a probability $$p$$ of passing the DC 13 Constitution saving throw, and we will assume that $$p$$ is greater than 0 and less than 1 (that is, passing and failing the save are both possible). Let us also, for simplicity, measure damage in number of d6’s (we can convert to actual damage at the end; 1d6 has a mean damage of 3.5).

Also note that the mean (average) of a discrete probability distribution, where getting $$x$$ has probability $$P(x)$$, is given by
$$sum_x x P(x),$$
where we sum over all values of $$x$$.

Before tackling the full problem, let us consider a subset of the problem. Suppose the victim has saved their way down to taking only 1d6 damage at the start of their turns. What is the average damage they will take then? We need to enumerate over each possible scenario. The victim could take 1d6 damage at the start of their turn then pass their save (with probability $$p$$) at the end of their turn, taking no further damage. The victim could fail one save then pass the second (with probability $$(1-p)p$$), taking 2d6 damage. And so on. The mean number of d6’s taken in the ‘1d6-per-round’ phase is
$$sum_{n=0}^infty (n+1) p (1-p)^n.$$

This is an arithmetico-geometric sequence. We can solve it with some basic algebraic manipulation.

$$sum_{n=0}^infty (n+1) p (1-p)^n = frac{p}{1-p} sum_{n=0}^infty (n+1) (1-p)^{n+1} = frac{p}{1-p} sum_{m=1}^infty m (1-p)^m = frac{p}{(1-p)} frac{(1-p)}{p^2} = frac{1}{p}.$$

In the 1d6-per-round phase, the victim takes an average of $$frac{1}{p}$$ lots of 1d6 damage (remember that $$p$$ is less than 1, so $$1/p$$ is greater than 1). It is simple to extrapolate that the 2d6-per-round phase will have a mean of $$displaystylefrac{2}{p}$$, and the 3d6-per-round phase will have a mean of $$displaystylefrac{3}{p}$$. Because the phases are independent, we are able to simply add together their means for the overall mean.

However, there is one tricky point with regards to the round in which the potion is consumed. If the victim drinks the potion on their turn, then the item text implies they get to make a save to reduce the damage before the start of their next turn. This means there is only a $$1-p$$ chance of entering the 3d6-per-round phase, so the mean damage taken from that phase is $$displaystylefrac{3(1-p)}{p}$$.

Let us bring everything together now. If the victim drinks the potion on their turn and fails their initial save, then they take an initial 3d6 damage plus $$frac{3(1-p)}{p}$$ d6 plus $$frac{2}{p}$$ d6 plus $$frac{1}{p}$$ d6 damage, for a total of
$$3+frac{3(1-p)}{p}+frac{2}{p}+frac{1}{p} = 3 + frac{3(2-p)}{p} = frac{6}{p} mbox{d6 damage,}$$
or $$displaystylefrac{21}{p}$$ damage.

If we make no assumptions about the initial save, then we get the above damage with probability $$1-p$$ and just 3d6 damage with probability $$p$$, for a net result of $$displaystylefrac{6(1-p)}{p} + 3p$$ d6 or $$10.5left(displaystylefrac{2(1-p)}{p} + pright)$$ damage.

If the victim ingests the poison outside their turn (such as with a Ready action, or by having another character administer it), then the mean damage from the 3d6-per-round phase is $$displaystylefrac{3}{p}$$ d6. If they fail their initial save, they take $$3 + displaystylefrac{6}{p}$$ d6 or $$10.5left(1+displaystylefrac{2}{p}right)$$ damage. If we make no assumptions about the initial save, they take $$frac{6}{p}-3$$ d6 or $$10.5left(displaystylefrac{2}{p}-1right)$$ damage.

Let us consider some concrete values of $$p$$. The average damage dealt by the potion of poison is (rounded to one decimal place)…

CON Save $$p$$ On turn, fail first save On turn Off turn, fail first save Off turn
-7 0.05 420 399.5 430.5 409.5
-6 0.10 210 190.1 220.5 119.5
-5 0.15 140 120.6 150.5 129.5
-4 0.20 105 86.1 115.5 94.5
-3 0.25 84 65.6 94.5 73.5
-2 0.30 70 52.2 80.5 59.5
-1 0.35 60 42.7 70.5 49.5
+0 0.40 52.5 35.7 63 42
+1 0.45 46.7 30.4 57.2 36.2
+2 0.50 42 26.3 52.5 31.5
+3 0.55 38.2 23.0 48.7 27.7
+4 0.60 35 20.3 45.5 24.5
+5 0.65 32.3 18.1 42.8 21.8
+6 0.70 30 16.4 40.5 19.5
+7 0.75 28 14.9 38.5 17.5
+8 0.80 26.3 13.7 36.8 15.8
+9 0.85 24.7 12.6 35.2 14.2
+10 0.90 23.3 11.8 33.8 12.8
+11 0.95 22.1 11.8 33.8 12.8
>11 1.00 10.5 10.5

## dnd 5e – What is the average damage dealt by a Potion of Poison?

Let us assume that our victim has a probability $$p$$ of passing the DC 13 Constitution saving throw, and we will assume that $$p$$ is greater than 0 and less than 1 (that is, passing and failing the save are both possible). Let us also, for simplicity, measure damage in number of d6’s (we can convert to actual damage at the end; 1d6 has a mean damage of 3.5).

Also note that the mean (average) of a discrete probability distribution, where getting $$x$$ has probability $$P(x)$$, is given by
$$sum_x x P(x),$$
where we sum over all values of $$x$$.

Before tackling the full problem, let us consider a subset of the problem. Suppose the victim has saved their way down to taking only 1d6 damage at the start of their turns. What is the average damage they will take then? We need to enumerate over each possible scenario. The victim could take 1d6 damage at the start of their turn then pass their save (with probability $$p$$) at the end of their turn, taking no further damage. The victim could fail one save then pass the second (with probability $$(1-p)p$$), taking 2d6 damage. And so on. The mean number of d6’s taken in the ‘1d6-per-round’ phase is
$$sum_{n=0}^infty (n+1) p (1-p)^n.$$

This is an arithmetico-geometric sequence. We can solve it with some basic algebraic manipulation.

$$sum_{n=0}^infty (n+1) p (1-p)^n = frac{p}{1-p} sum_{n=0}^infty (n+1) (1-p)^{n+1} = frac{p}{1-p} sum_{m=1}^infty m (1-p)^m = frac{p}{(1-p)} frac{(1-p)}{p^2} = frac{1}{p}.$$

In the 1d6-per-round phase, the victim takes an average of $$1/p$$ lots of 1d6 damage (remember that $$p$$ is less than 1, so $$1/p$$ is greater than 1). It is simple to extrapolate that the 2d6-per-round phase will have a mean of $$2/p$$, and the 3d6-per-round phase will have a mean of $$3/p$$. Because the phases are independent, we are able to simply add together their means for the overall mean.

However, there is one tricky point with regards to the round in which the potion is consumed. If the victim drinks the potion on their turn, then the item text implies they get to make a save to reduce the damage before the start of their next turn. This means there is only a $$1-p$$ chance of entering the 3d6-per-round phase, so the mean damage taken from that phase is $$3(1-p)/p$$.

Let us bring everything together now. If the victim drinks the potion on their turn and fails their initial save, then they take an initial 3d6 damage plus $$3(1-p)/p$$ d6 plus $$2/p$$ d6 plus $$1/p$$ d6 damage, for a total of $$3+3(1-p)/p+2/p+1/p = 3 + 3(2-p)/p = 6/p$$ d6 damage, or $$21/p$$ damage.

If we make no assumptions about the initial save, then we get the above damage with probability $$1-p$$ and just 3d6 damage with probability $$p$$, for a net result of $$6(1-p)/p + 3p$$ d6 or $$10.5(2(1-p)/p + p)$$ damage.

If the victim ingests the poison outside their turn (such as with a Ready action, or by having another character administer it), then the mean damage from the 3d6-per-round phase is $$3/p$$ d6. If they fail their initial save, they take $$3 + 6/p$$ d6 or $$10.5(1+2/p)$$ damage. If we make no assumptions about the initial save, they take $$6/p-3$$ d6 or $$10.5(2/p-1)$$ damage.

Let us consider some concrete values of $$p$$. The average damage dealt by the potion of poison is (rounded to one decimal place)…

CON Save $$p$$ On turn, fail first save On turn Off turn, fail first save Off turn
-7 0.05 420 399.5 430.5 409.5
-6 0.10 210 190.1 220.5 119.5
-5 0.15 140 120.6 150.5 129.5
-4 0.20 105 86.1 115.5 94.5
-3 0.25 84 65.6 94.5 73.5
-2 0.30 70 52.2 80.5 59.5
-1 0.35 60 42.7 70.5 49.5
+0 0.40 52.5 35.7 63 42
+1 0.45 46.7 30.4 57.2 36.2
+2 0.50 42 26.3 52.5 31.5
+3 0.55 38.2 23.0 48.7 27.7
+4 0.60 35 20.3 45.5 24.5
+5 0.65 32.3 18.1 42.8 21.8
+6 0.70 30 16.4 40.5 19.5
+7 0.75 28 14.9 38.5 17.5
+8 0.80 26.3 13.7 36.8 15.8
+9 0.85 24.7 12.6 35.2 14.2
+10 0.90 23.3 11.8 33.8 12.8
+11 0.95 22.1 11.8 33.8 12.8
>11 1.00 10.5 10.5

## calculus – Calculate average revenue per user over time with cancellation rates factored in

I’m hoping the best brains can help me out here with a equation that calculates how much revenue I can expect per user over time.

The input values are as follows.
I provide a service that have $$1000$$ users that pays $$10$$ per month. Think similar to Spotify.
Each month $$3$$ percent or $$30$$ of the users cancels the service and stops paying. But each month $$30$$ new users sign up and pays me $$10$$ so the net income and user base is the same.
If a customer have been part of the service for $$60$$ months, it will automatically cancel.

Out of the percent that cancels, some will have been a paying customer for only $$1$$ month, meaning the revenue of that user is $$10cdot 1$$, and some would have been a customer for $$60$$ months, so their revenue is going to be $$10 cdot 60$$.

I’m looking for an equation that will simulate on average how many months these $$30$$ customers that sign up will be a paying customer, basically how many months of revenue can I expect per user given the input values above. Like for example on average each customer that signs up this month will stay for $$27$$ months, so the revenue will be $$27cdot 10$$.

As a next step I wanna see different outcomes if cancellation rates goes to 2 or 4 percent and if new customer sign up per month is higher than cancellations, but as a first step getting the basic equation in place.

Thank you so much for any help!