What does it mean by independent trials? Can you give a specific example. Surely, if the probability is fixed than the trials are independent. I’m confused.

Original post by maggiehodgson

What does it mean by independent trials? Can you give a specific example. Surely, if the probability is fixed than the trials are independent. I’m confused.

Independence means the probability for this trial does not depend on whatt happened before, so throwing a die is a simple/obvious one. Independence means the joint probability is the product of the (constant) probabilities of each trial so the "^n" part for a binomial and as the trials are identical, the probability is unchanged. So p(AnB) = p(A)*p(B).

A simple dependent one is drawing balls from a bag, without replacement. You can get the joint probability by multiplying probabilities together (multiply along a branch in a tree), though those probabilities are really conditional probabilities, so they depend on what has happened (higher up in the tree - which balls have been drawn beforehand). So p(AnB) = p(A)*p(B|A).

You could imagine independence as p(B)=p(B|A), so the probability is fixed, as you seem to be implying. So knowlege about what has happened before has no effect on the current trial. Binomial is iid so independent and identially distributed.

(edited 1 month ago)

Original post by maggiehodgson

What does it mean by independent trials? Can you give a specific example. Surely, if the probability is fixed than the trials are independent. I’m confused.

Suppose I have a black ball and a white ball. I pick one at random and put it in a bag.

Each trial X_n corresponds to picking the ball out of the bag, noting its colour, and replacing it.

Then each X_n is identically distributed with p(black)=p(white)=1/2.

But obviously all the X_n are actually equal. If you know the result of X_1 you know the result of all the subsequent trials.

So not at all independent.

Original post by mqb2766

Independence means the probability for this trial does not depend on whatt happened before, so throwing a die is a simple/obvious one. Independence means the joint probability is the product of the (constant) probabilities of each trial so the "^n" part for a binomial and as the trials are identical, the probability is unchanged. So p(AnB) = p(A)*p(B).

A simple dependent one is drawing balls from a bag, without replacement. You can get the joint probability by multiplying probabilities together (multiply along a branch in a tree), though those probabilities are really conditional probabilities, so they depend on what has happened (higher up in the tree - which balls have been drawn beforehand). So p(AnB) = p(A)*p(B|A).

You could imagine independence as p(B)=p(B|A), so the probability is fixed, as you seem to be implying. So knowlege about what has happened before has no effect on the current trial. Binomial is iid so independent and identially distributed.

A simple dependent one is drawing balls from a bag, without replacement. You can get the joint probability by multiplying probabilities together (multiply along a branch in a tree), though those probabilities are really conditional probabilities, so they depend on what has happened (higher up in the tree - which balls have been drawn beforehand). So p(AnB) = p(A)*p(B|A).

You could imagine independence as p(B)=p(B|A), so the probability is fixed, as you seem to be implying. So knowlege about what has happened before has no effect on the current trial. Binomial is iid so independent and identially distributed.

But in that case the probabilities are changing from selection to selection. I’m asking if probability of a success remains constant what could make events not independent. The binomial distribution requires both a fixed probability and independence. So, if the probability of success does not change from event to event can you give me an example when the events are not independent. Thanks

Original post by maggiehodgson

But in that case the probabilities are changing from selection to selection. I’m asking if probability of a success remains constant what could make events not independent. The binomial distribution requires both a fixed probability and independence. So, if the probability of success does not change from event to event can you give me an example when the events are not independent. Thanks

Its obviously not binomial, but if you had something like the probability of being ill each day is 0.01. However, being ill on two successive days is not 0.01^2 as the conditional probability of being ill on the second day is ~0.7. given youre ill on the first day.

Obv made up numbers in the previous example

(edited 1 month ago)

Original post by maggiehodgson

But in that case the probabilities are changing from selection to selection. I’m asking if probability of a success remains constant what could make events not independent. The binomial distribution requires both a fixed probability and independence. So, if the probability of success does not change from event to event can you give me an example when the events are not independent. Thanks

My post is an example where the probability is constant from trial to trial but the values are not independent.

Here's another interesting example. Fix N >= 2.

For each trial n, pick a random number X between 0 and 2^N - 1. Count the number of 1's in its binary representation. If this is even, add 2^N to X. Then define random variables X_0, X_1, ..., X_N such that X_k is the binary digit corresponding to 2^k in the binary representation of X.

E.g. Set N = 3, pick a random number between 0 and 7. For example 5. This is 101 in binary, so has an even number of 1s, so add 2^3 = 8 to get 13, or 1101 in binary. Then X_0 = 1, X_1 = 0, X_2 = 1, X_3 = 1 (just read from the binary representation of 13 in reverse order).

X_0, X_1, ..., X_N are N+1 different random variables.

It is not difficult to show the following:

Each X_i = 1 with probability of 1/2, and = 0 with probability of 1/2.

Any group of less than N+1 items are independent. (loosely speaking, if you pick any N of the X_i from X_0,X_1,...,X_N, you can't discern any relationship between them, no matter how many trials you run).

But the N+1 items together are not independent. By construction, they must always add up to an odd number, so given any N items, you can predict the N+1th item with 100% accuracy.

(edited 1 month ago)

Original post by mqb2766

Its obviously not binomial, but if you had something like the probability of being ill each day is 0.01. However, being ill on two successive days is not 0.01^2 as the conditional probability of being ill on the second day is ~0.7. given youre ill on the first day.

Obv made up numbers in the previous example

Obv made up numbers in the previous example

Thanks. That’s a great example

Original post by mqb2766

Its obviously not binomial, but if you had something like the probability of being ill each day is 0.01. However, being ill on two successive days is not 0.01^2 as the conditional probability of being ill on the second day is ~0.7. given youre ill on the first day.

Obv made up numbers in the previous example

Obv made up numbers in the previous example

Wonderful. That makes so much sense

Original post by maggiehodgson

Wonderful. That makes so much sense

Along the lines of Dfranklins examples, another one that came to mind afterwards was based on the sum of two dice. For two 2 sided dice with values 1 and 2 (wont roll very well), then the sum would take values 2,3,4 and if theyre fair and independent (or the game below is run over a long time), then the probabilities would be 1/4,1/2,1/4 as usual.

Here the game is that after a roll, one of the dice is retained (randomly selection) and the other rolled again to get the new sum. So if the previous sum was 4 say, then both dice would have been 2 and the retained die must be 2. So the new sum(based on the roll of a single die) could only be 3 or 4 with probabilities 1/2 and 1/2. So given the previous sum, the new sum probabilities can be affected so dependent. Given that the previous sum is 2, you have an analogous case for new scores of 2 and 3 with probabiltiies 1/2,1/2. However, given that the previous sum is 3, you have the usual case in the previous paragraph as the retained die could be either 1 or 2 and similarly for the new rolled die, so the score could be 2,3 or 4 with probabilities 1//4,1/2,1/4

It would be easy enough to draw as the usual probability tree(s), but to cast in binomial language then win is scoring 4 and p=p(4)=1/4. However, p(4|2)=0, p(4|3)=1/4 and p(4|4)=1/2.

(edited 1 month ago)

Original post by maggiehodgsonWhat does it mean by independent trials? Can you give a specific example. Surely, if the probability is fixed than the trials are independent. I’m confused.

probability never changes (therefore u always use binomial dist)

Original post by alaziziomar_

probability never changes (therefore u always use binomial dist)

No.

Original post by DFranklin

No.

Can you correct me please?

Thank you for pointing it out

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