I want to know is there a function to calculate the inverse cdf of poisson distribution? So that I can use inverse CDF of poisson to generate a set of poisson distributed random number.
A) Inverse CDF of Poisson distribution
The inverse CDF at q is also referred to as the q quantile of a distribution. For a discrete distribution distribution . the inverse CDF at q is the smallest integer x such that CDF[dist,x]≥q.. The Poisson distribution is a discrete distribution that models the number of events based on a constant rate of occurrence. The Poisson distribution can be used as an approximation to the binomial when the number of independent trials is large and the probability of success is small. A common application of the Poisson distribution is predicting the number of events over a specific time, such as the number of cars arriving at a toll plaza in 1 minute.
Formula
The probability mass function (PMF) is:
mean = λ
variance = λ
Notation
Term Description
e base of the natural logarithm
Reference: Methods and Formulas for Inverse Cumulative Distribution Functions
B) Excel Function: Excel provides the following function for the Poisson distribution:
POISSON(x, μ, cum)
where μ = the mean of the distribution and cum takes the values TRUE and FALSE
POISSON(x, μ, FALSE) = probability density function value f(x) at the value x for the Poisson distribution with mean μ.
POISSON(x, μ, TRUE)= cumulative probability distribution function F(x) at the value x for the Poisson distribution with mean μ.
Excel 2010/2013/2016 provide the additional function POISSON.DIST which is equivalent to POISSON.
Reference: Office Support POISSON.DIST Function
C) Excel doesn’t provide a worksheet function for the inverse of the Poisson distribution.
Instead you can use the following function provided by the Real Statistics Resource Pack. It’s a free download for Excel various versions.
POISSON_INV(p, μ) = smallest integer x such that POISSON(x, μ, TRUE) ≥ p
Note that the maximum value of x is 1,024,000,000. A value higher than this indicates an error.
Reference: Real Statistics Using Excel
D)
Reference to MREXCEL.COM web site a query related to your question quoted below seems to be related to your question.
Not sure if anyone can help with this. Basically I'm trying to find out how to apply the reverse of the Poisson function in excel. So as of now I have poisson(x value, mean, true-cumulative) and that lets me get the probability for that occurence. Basically I want to know how I can get the minimum/maximum x value based on a given probability.
So if I have a list of data (700 rows) and I want to find out what the minimum starting value should be given a desired average and the fact that I want the lowest value to be at the 0.05% probability. So 0.05% = (x, 35, True) solve for x. I know I can prob do this with solver, but I am trying to figure out a way to do this formulaicly without having to use the solver (as I may have to use this many times).
The code referred to here covers the inverse of the poisson formula when using True in the excel formula. It does not cover the inverse of the poisson formula when using False in the excel formula.
Re: Reverse Poisson?
Originally Posted by shg
A further mod to accommodate large means:
Code:
Function PoissonInv(Prob As Double, Mean As Double) As Variant
' shg 2011, 2012, 2014, 2015-0415
' For a Poisson process with mean Mean, returns a three-element array:
' o The smallest integer N such that POISSON(N, Mean, True) >= Prob
' o The CDF for N-1 (which is < Prob)
' o The CDF for N (which is >= Prob)
-------Reference :> https://www.mrexcel.com/forum/excel-questions/507508-reverse-poisson-2.html>
E) Why doesn't Excel have a POISSON.INV function?
Discussion on Referred web page have references to some formulas for calculating related information desired by OP.
You could use the following.
With the Poisson mean named lambda, enter the following in an newly inserted worksheet.
A1: =IF(ROWS(A$1:A1)<=4*lambda,POISSON(ROWS(A$1:A1)-1,lambda,1))
Fill A1 down into A2:A1000 (4 times as many rows as your most typical lambda value). Name the A1:A1000 range POISSON.CDF. Then use the formula
=MATCH(n,POISSON.CDF)-1
to give the results a POISSON.INV(n,lambda) function would.
If you want this for varying lambda, use the array formula
=MATCH(n,POISSON(ROW($A$1:INDEX($A:$A,4*lambda+1),lambda,1))-1
Reference Shared Link
Hope That Helps.
=MATCH(RAND(),MMULT((ROW(INDIRECT(ADDRESS(1,1)&":"&ADDRESS(MAX(lambda,5+lambda* 45/50)+6* SQRT(lambda)+3,1)))=COLUMN(INDIRECT(ADDRESS(1,1)&":"&ADDRESS(1,MAX(lambda,5+lambda* 45/50)+6* SQRT(lambda)+2))))+0,MMULT((ROW(INDIRECT(ADDRESS(1,1)&":"&ADDRESS(MAX(lambda,5+lambda* 45/50)+6* SQRT(lambda)+2,1)))=(COLUMN(INDIRECT(ADDRESS(1,1)&":"&ADDRESS(1,MAX(lambda,5+lambda* 45/50)+6* SQRT(lambda)+1)))+1))+0,POISSON(ROW($A$1:INDEX($A:$A,MAX(lambda,5+lambda* 45/50)+6* SQRT(lambda)+1))-1,lambda,1)))+(ROW(INDIRECT(ADDRESS(1,1)&":"&ADDRESS(MAX(lambda,5+lambda* 45/50)+6* SQRT(lambda)+3,1)))=(COLUMN(INDIRECT(ADDRESS(1,1)&":"&ADDRESS(1,1)))+FLOOR(MAX(lambda,5+lambda* 45/50)+6* SQRT(lambda)+2,1)))+0)-1
It is quite slow for lambda >1000.
This expands on the array formula
=MATCH(C4,POISSON(ROW($A$1:INDEX($A:$A,4*lambda+1)),lambda,1))-1
shared above by skkakkar, by prepending the array with 0 and appending with 1, following Is there a way to concatenate two arrays in Excel without VBA? .
The rest is mostly making the array shorter by replacing 4* lambda with 6* SQRT(lambda).
Related
Given an NxM array of positive integers, how would one go about selecting integers so that the maximum sum of values is achieved where there is a maximum of x selections in each row and y selections in each column. This is an abstraction of a problem I am trying to face in making NCAA swimming lineups. Each swimmer has a time in every event that can be converted to an integer using the USA Swimming Power Points Calculator the higher the better. Once you convert those times, I want to assign no more than 3 swimmers per event, and no more than 3 races per swimmer such that the total sum of power scores is maximized. I think this is similar to the Weapon-targeting assignment problem but that problem allows a weapon type to attack the same target more than once (in my case allowing a single swimmer to race the same event twice) and that does not work for my use case. Does anybody know what this variation on the wta problem is called, and if so do you know of any solutions or resources I could look to?
Here is a mathematical model:
Data
Let a[i,j] be the data matrix
and
x: max number of selected cells in each row
y: max number of selected cells in each column
(Note: this is a bit unusual: we normally reserve the names x and y for variables. These conventions can help with readability).
Variables
δ[i,j] ∈ {0,1} are binary variables indicating if cell (i,j) is selected.
Optimization Model
max sum((i,j), a[i,j]*δ[i,j])
sum(j,δ[i,j]) ≤ x ∀i
sum(i,δ[i,j]) ≤ y ∀j
δ[i,j] ∈ {0,1}
This can be fed into any MIP solver.
Introduction
I have written code to give me a set of numbers in '36 by q' format ( 1<= q <= 36), subject to following conditions:
Each row must use numbers from 1 to 36.
No number must repeat itself in a column.
Method
The first row is generated randomly. Each number in the coming row is checked for the above conditions. If a number fails to satisfy one of the given conditions, it doesn't get picked again fot that specific place in that specific row. If it runs out of acceptable values, it starts over again.
Problem
Unlike for low q values (say 15 which takes less than a second to compute), the main objective is q=36. It has been more than 24hrs since it started to run for q=36 on my PC.
Questions
Can I predict the time required by it using the data I have from lower q values? How?
Is there any better algorithm to perform this in less time?
How can I calculate the average number of cycles it requires? (using combinatorics or otherwise).
Can I predict the time required by it using the data I have from lower q values? How?
Usually, you should be able to determine the running time of your algorithm in terms of input. Refer to big O notation.
If I understood your question correctly, you shouldn't spend hours computing a 36x36 matrix satisfying your conditions. Most probably you are stuck in the infinite loop or something. It would be more clear of you could share code snippet.
Is there any better algorithm to perform this in less time?
Well, I tried to do what you described and it works in O(q) (assuming that number of rows is constant).
import random
def rotate(arr):
return arr[-1:] + arr[:-1]
y = set([i for i in range(1, 37)])
n = 36
q = 36
res = []
i = 0
while i < n:
x = []
for j in range(q):
if y:
el = random.choice(list(y))
y.remove(el)
x.append(el)
res.append(x)
for j in range(q-1):
x = rotate(x)
res.append(x)
i += 1
i += 1
Basically, I choose random numbers from the set of {1..36} for the i+q th row, then rotate the row q times and assigned these rotated rows to the next q rows.
This guarantees both conditions you have mentioned.
How can I calculate the average number of cycles it requires?( Using combinatorics or otherwise).
I you cannot calculate the computation time in terms of input (code is too complex), then fitting to curve seems to be right.
Or you could create an ML model with iterations as data and time for each iteration as label and perform linear regression. But that seems to be overkill in your example.
Graph q vs time
Fit a curve,
Extrapolate to q = 36.
You might want to also graph q vs log(time) as that may give an easier fitted curve.
I’ve used Excel in the past but the calculations including the Poisson-Distribution took a while, that’s why I switched to SQL. Soon I’ve recognized that SQL might not be a proper solution to deal with statistical issues. Finally I’ve decided to switch to Matlab but I’m not used to it at all, my problem Is the following:
I’ve imported a .csv-table and have two columns with values, let’s say A and B (110 x 1 double)
These values both are the input values for my Poisson-calculations. Since I wanna calculate for at least the first 20 events, I’ve created a variable z=1:20.
When I now calculated let’s say
New = Poisspdf(z,A),
it says something like non-scalar arguments must match in size.
Z only has 20 records but A and l both have 110 records. So I’ve expanded Z= 1:110 and transposed it:
Znew = Z.
When I now try to execute the actual calculation:
Results = Poisspdf(Znew,A).*Poisspdf(Znew,B)
I always get only a 100x1 Vector but what I want is a matrix that is 20x20 for each record of A and B (based on my actual choice of z=1:20, I only changed to z=1:110 because Matlab told that they need to match in size).
So in this 20x20 Matrix there should always be in each cell the result of a slightly different calculation (Poisspdf(Znew,A).*Poisspdf(Znew,B)).
For example in the first cell (1,1) I want to have the result of
Poisspdf(0,value of A).*Poisspdf(0,value of B),
in cell(1,2): Poisspdf(0,value of A).*Poisspdf(1,value of B),
in cell(2,1): Poisspdf(1,value of A).*Poisspdf(0,value of B),
and so on...assuming that it’s in the Format cell(row, column)
Finally I want to sum up certain parts of each 20x20 matrix and show the result of the summed up parts in new columns.
Is there anybody able to help? Many thanks!
EDIT:
Poisson Matrix in Excel
In Excel there is Poisson-function: POISSON(x, μ, FALSE) = probability density function value f(x) at the value x for the Poisson distribution with mean μ.
In e.g. cell AD313 in the table above there is the following calculation:
=POISSON(0;first value of A;FALSE)*POISSON(0;first value of B;FALSE)
, in cell AD314
=POISSON(1;first value of A;FALSE)*POISSON(0;first value of B;FALSE)
, in cell AE313
=POISSON(0;first value of A;FALSE)*POISSON(1;first value of B;FALSE)
, and so on.
I am not sure if I completely understand your question. I wrote this code that might help you:
clear; clc
% These are the lambdas parameters for the Poisson distribution
lambdaA = 100;
lambdaB = 200;
% Generating Poisson data here
A = poissrnd(lambdaA,110,1);
B = poissrnd(lambdaB,110,1);
% Get the first 20 samples
zA = A(1:20);
zB = B(1:20);
% Perform the calculation
results = repmat(poisspdf(zA,lambdaA),1,20) .* repmat(poisspdf(zB,lambdaB)',20,1);
% Sum
sumFinal = sum(results,2);
Let me know if this is what you were trying to do.
I use datas in excel to produce a graphic.
Then I make a regression, and have an equation. I'd like to know what value would be obtained from the regression (for example, x = 7,6 is the value for which I wanna know an estimation of y).
It is an approximation with a 6 degree polynome.
One wimple method would be this : I have the equation, so I could use it
However, I wondered if there is a fast method to do it? Like I enter 7,6 somewhere to have the result quickly?
if you are looking at a linear regression line (straight line) you could try the forecast formula
=forecast(X, Known Ys, Known Xs)
you could also build your own equation automatically from
=linest(...)
I found the following on a site describing the capabilities of the linest function in excel:
In addition to using LOGEST to calculate statistics for other
regression types, you can use LINEST to calculate a range of other
regression types by entering functions of the x and y variables as the
x and y series for LINEST. For example, the following formula:
=LINEST(yvalues, xvalues^COLUMN($A:$C))
works when you have a single column of y-values and a single column of
x-values to calculate the cubic (polynomial of order 3) approximation
of the form:
y = m1*x + m2*x^2 + m3*x^3 + b
You can adjust this formula to calculate other types of regression,
but in some cases it requires the adjustment of the output values and
other statistics.
or look at:
=trend
I'm trying to use NORMDIST function in Excel to create a bell curve, but the output is strange.
My mean is 0,0000583 and standard deviation is 0,0100323 so when I plug this to the function NORMDIST(0,0000583; 0,0000583; 0,0100323; FALSE) I expect to get something close to 0,5 as I'm using the same value as the mean probability of this value should be 50%, but the function gives an output of 39,77 which is clearly not correct.
Why is it like this?
A probability cannot have values greater than 1, but a density can.
The integral of the entire range of a density function is equal 1, but it can have values greater than one in specific interval. Example, a uniform distribution on the interval [0, ½] has probability density f(x) = 2 for 0 ≤ x ≤ ½ and f(x) = 0 elsewhere. See below:
=NORMDIST(x, mean, dev, FALSE) returns the density function. Densities are probabilities per unit. It is almost the probability of a point, but with a very tiny range interval (the derivative in the point).
shg's answer here, explain how to get a probability on a given interval with NORMIDIST and also in what occasions it can return a density greater than 1.
For a continuous variable, the probability of any particular value is zero, because there are an infinite number of values.
If you want to know the probability that a continuous random variable with a normal distribution falls in the range of a to b, use:
=NORMDIST(b, mean, dev, TRUE) - NORMDIST(a, mean, dev, TRUE)
The peak value of the density function occurs at the mean (i.e., =NORMDIST(mean, mean, dev, FALSE) ), and the value is:
=1/(SQRT(2*PI())*dev)
The peak value will exceed 1 when the deviation is less than 1 / sqrt(2pi) ~ 0.399,
which was your case.
This is an amazing answer on Cross Validated Stack Exchange (statistics) from a moderator (whuber), that addresses this issue very thoughtfully.
It is returning the probability density function whereas I think you want the cumulative distribution function (so try TRUE in place of FALSE) ref.