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I toss a die five times X = 1, 2, 3, 4, 5. Means there are five throws. On each throw, we get a number on the die. This means we have 5 values for 5 throws.
Y = 1, 5, 4, 3, 2

On the first throw (X = 1), I get Y = 1 on the die. In the second throw (X = 2), I get a value of Y=5 on the die.

I compute the data proportions by dividing each value by the total sum of the values. Normalized data : 1/15, 5/15, 4/15, 3/15, 2/15
Here, 15 is the sum of all the values that I have observed during all of my 5 throws.

Now, I know that although the sum of the proportional values will be one, we cannot call this a probability distribution of Y given X. Is there any way to link the proportion values with the probability distribution of Y? I wish to create a probability distribution which would reflect the proportion of Y for each X.

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    $\begingroup$ What are X an Y? And if you toss it 5 times, why do you divide by 15? $\endgroup$
    – gunes
    May 23, 2023 at 6:15
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    $\begingroup$ Are the Y's intended to be counts within each X category? $\endgroup$
    – Glen_b
    May 23, 2023 at 7:04
  • $\begingroup$ Apologies for the confusion. X is the trial number. I throw the die 5 times, so X is 5. Y is the value that I observed on the die after I throw it. $\endgroup$
    – nivedita
    May 23, 2023 at 9:46
  • $\begingroup$ If 6 is possible, but not observed in the sample, that's part of the data. So you have observed proportions 1/5 5 times and 0/5 times. It's a matter of statistics as much as probability that you don't have a big enough sample to do anything much with. Even in principle not all possible outcomes can be observed in the same sample, so that at least one observed proportion must be 0. $\endgroup$
    – Nick Cox
    May 23, 2023 at 9:52

1 Answer 1

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I'm assuming that $X$ are the values (e.g. $X=3$ means throwing ⚂) and $Y$ are the counts (so $X=3$ and $Y=4$ means that you observed ⚂ four times out of $\sum_j Y_j = 15$ throws in total).

First of all, your language is confusing. The "probability distribution of Y given X" sounds like a conditional probability, while it doesn't seem that you are conditioning on anything here, but rather counting the outcomes.

As for your question, $\hat p_i = Y_i / \sum_j Y_j$ would be the estimate of the probability of observing the $X_i$ outcome using the empirical probability. This would be also the maximum likelihood estimator for the categorical distribution that the dice throws are following. Alternatively, you can use the Bayesian estimator if you can define a prior for the probabilities. It is a valid way of estimating the distribution, though the obvious limitation is that the less data you have, the less precise and trustworthy the result would be.

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