Theory and Experiment of intercepting some samples conforming to exponential distribution

Recently, when solving the problem of processing samples that conform to the exponential distribution, we made a hypothesis. Then we need to make a small experiment to confirm the correctness of the theory simply derived based on the hypothesis.

First, assume that given a sample set with N as the total number, the elements in the sample set conform to the exponential distribution, that is, the value of X of each element in the sample set S conforms to the exponential distribution X ~ with the lambda parameter ~ Exp (lambda). If I specify another length of N to intercept all the sample elements, that is, all the elements whose X is less than or equal to n.

Question: 1) How many such elements are represented by N0? 2) What is the sum of all the intercepted elements, represented by L?

A. Simple Derivation:

1) the first small problem is my idea: First, find the cumulative distribution probability F (n, lambda) with element x not greater than N ), then, the number of all elements not greater than N is the embodiment of the total number of samples in F (n, lambda. That is

2) The second question is: first obtain the expected E (x <= N) of all elements less than or equal to N, and then l is the expected E (x <= N) overall sample. That is

F (x) is the probability density function of exponential distribution.

B. Next, use an experiment for verification.

1) Code; 2) effects; 3) conclusion.

1)

clear% ---- 1) generate S with Exprnd()S = [];cnt_elements = 1e6;Mu = 5;for i=1:cnt_elementsS(i)= exprnd(Mu);end% ---- 2) countingn_threshold = 3;selected_elements_idx = [];selected_elements_idx = find(S <= n_threshold);% -- a. count of selected elements within threshold.CNT_selected = size(selected_elements_idx);% -- b. sum of the selected elements.sum_sel_ones = sum( S(selected_elements_idx) );% ----- 3) analysis of N0:lam = 1.0/Mu;n = n_threshold;N = cnt_elements;N0 = N * (1-exp(-1*lam*n) );% ----- 4) analysis of L.L = (N/lam)*(1 - (lam*n + 1)/exp(lam*n));% ----- 5) Compare CNT_selected with N0.CNT_selectedN0% ----- 6) Compare sum_sel_ones with L.sum_sel_onesL

2) output:

Cnt_selected =
1 451172

N0 =
4.20.9e + 005

Sum_sel_ones =
6.0934e + 005

L =
6.0951e + 005

3) the output shows that the experiment results are roughly in line with the theoretical derivation.

Davy_h

2014-10-15