# Thread: How to prove the following distribution

1. ## How to prove the following distribution

Hi,

I have the following problem I would like to get some input for. Let say I have a set of numbers from 1 to 500 : $A=\{1..500\}$ and i am preforming the following experiment (sampling with replacement). I randomly pick 100 numbers from A. Since i am assuming the picking is done randomly i can assume that each pick happens with probability $\frac{1}{500}$ that is i have an equal chance to pick 1 and 40. since i am picking 100 numbers from those 500 (with replacement) i can get for a result 100 1's each with the same probability $\frac{1}{500}$. eventually if i repeat this experiment many times i'll get two 1's two 2's, two 3's and so on... (is this clear so far). Now a single experiment is picking 100 numbers from those 500 allowing replacement. since in one experiment i can get four 1's and in the next (assuming i repeated the experiment) i can get ten 1's my question is will the frequencies of 1's be also uniformly distributed after i repeat the experiment many times. And how to prove that? If not how can i prove the opposite and figure out the distribution.

b

2. ## Re: How to prove the following distribution

The frequency of 1's will always be the same no matter how many 1's you go in the previous experiments. This comes from one fact: the events are independent of each other. Since you are replacing the numbers each time the next number you draw cannot be affected by the previous numbers you draw, so all the previous numbers from previous experiments cannot change the frequency of the next number.

3. ## Re: How to prove the following distribution

Thank you. This makes sence. If I may, i have a follow up question. Let say that instead of repeating the experiment many times i repeat the experiment 1000 times in that case what would be the average frequency of 1's? So what confuses me is the way the experiment is carried out. I am always picking 100 random numbers out of 500. Repeating this 1000 times i would expect that every number gets picked on average 2 times (right??) However if i look at the frequencies of any particular number let say 1, than the probability of getting two 1's or fourty 1's is the same and is the function of the number of random numbers i have picked in each experiment, which is $\frac{1}{100}$. I know that my average is $\sum_{i=1}^{100} x \times P(x)$ where x is the frequency of 1's (or any other value) and $P(x) = \frac{1}{100}$. But how do i connect that with repeating the experiment 1000 time. So given that i repeat the experiment 1000 times what will be average frequency of 1's?

thank you once again

b

4. ## Re: How to prove the following distribution

If you take a sample 1000 times from the 500 numbers then you would expect to get the number 1 twice. But because of the variability the number of 1's there is a high chance that you wont get exactly 2.

Your formula for the average is the average value of the number you pick, which is 250.5 (half way between 1 and 500). This is not the average for the number of 1's you will get.
And $P(x)=\frac{1}{500}$

The number of 1's you get follows a binomial distribution, have you studied this?

From 1000 samples the probability for getting a 1 once is calculated as:

$\text{[Probability of getting a 1]}\times \text{[Probability of not getting a 1 999 times]} \times \text{[number of ways of picking one 1 from 1000 samples]}$

The probability of getting a 1 from one sample is $\frac{1}{500}$

The probability of not getting a 1 from one sample is $\frac{499}{500}$

The probability of not getting a 1 999 times is $\Big( \frac{499}{500} \Big)^{999}$

When you get one 1 from 1000 attempts it could be on the first sample, the second sample, all the way up to the 1000th sample. So there are 1000 ways in which you can sample one 1 from 1000 attempts.

Putting these into the formula from before

$\text{[Probability of getting a 1]}\times \text{[Probability of not getting a 1 999 times]} \times \text{[number of ways of picking one 1 from 1000 samples]}$

$\frac{1}{500}\times \Big(\frac{499}{500} \Big)^{999} \times 1000= 0.27067$

Likewise, the chance of getting a 1 twice is calculated as:

$\text{[Probability of getting a 1 twice]}\times \text{[Probability of not getting a 1 998 times]} \times \text{[number of ways of picking two 1's from 1000 samples]}$

$\Big(\frac{1}{500} \Big)^2 \times \Big(\frac{499}{500} \Big)^{998} \times 499500= 0.27094$

The probabilities of getting one 1 and getting two 1's are similar but not exactly the same. If you were to calculate the probability of getting three, four, five 1's then you would see that very quickly the probability becomes very small
If you want to know why the number of ways of picking two 1's from 1000 samples is 499500 I suggest you read a bit more about the binomial distribution.

5. ## Re: How to prove the following distribution

Thank you so much for helping me with this. Please don't stop. I am not quite sure i understand what you are saying. Yes, I understand, that if i sample 1000 times from 500 i could roughly expect two. But what I am doing in my experiment is, I am sampling 1000 times a set of 100 numbers from the set of 500 numbers. In a single sampling of 100 numbers i can get one hundred 1's. but also i can get only two 1's. If i get two 1's in my first sampling I would count : i got two 1's, one time. in the next iteration again i pick 100 numbers from 500 and again i get two 1's so i increment the frequency of two 1's from one to two. I repeat this type of sampling 1000 times and my question is what will be the most frequent frequency of 1's. where the most frequent frequency of 1's is the mean of frequencies of 1's. So it is a two dimensional problem. Maybe you understood me the first time but i am having a hard time connecting your solution to my problem. Could you please spare some more time in helping me with this problem. Thank you so much .

b

6. ## Re: How to prove the following distribution

Well if you are wondering about what the most common number of 1's from 100 samples is then you want to know the mode.
The most common number of 1's to observe from 100 samples where the chance of getting a 1 from one sample is $\frac{1}{500}$ can be calculated as:
$\text{round down(}(100+1)\frac{1}{500} \text{)} = \text{round down(}0.202 \text{)}=0$

The chance of getting zero 1's is 81.9%

my question is what will be the most frequent frequency of 1's. where the most frequent frequency of 1's is the mean of frequencies of 1's.
I think you meant the mode, but if you did mean the mean...
When you take 100 samples the number of 1's you would expect to get is $100\times \frac{1}{500}=0.2$. This is the mean, and when you repeat the experiment 1000 times you would have something similar to an average of 0.2 1's per 100 samples.

7. ## Re: How to prove the following distribution

Hi, I know it is a little bit late to continuet the thread but i studied up a bit on basic statistics and probability and i now understand what you were talking about. And yes i want the mode not the mean. However, to clarify one more thing. If wanded the mean and the mean is 0.2 and i am only dealing with discrete values and i wanted to count the number of times i got zero 1's in my experiment that was now repeated 100000 times (100000 times i sampled 100 numbers out of 500 - with replacement) then i need to multiply the probability of getting zero 1's (0.819) with what ???

thnx

8. ## Re: How to prove the following distribution

To find the expected number of times you would get zero 1's in your experiment, multiply the probability of getting zero 1's by the number of samples you will take.

$0.819\times 100000\times 100= 8190000$

9. ## Re: How to prove the following distribution

Thank you very much, it all makes sense now