Let $X_1, X_2$ be two discrete random variables. Each random variable takes two values: $1, 2$

The probability distribution $p_1$ over $X_1, X_2$ is given by

$$p_1(X_1=1, X_2 = 1) = \dfrac{1}{4}$$ $$p_1(X_1=1, X_2 = 2) = \dfrac{1}{4}$$ $$p_1(X_1=2, X_2 = 1) = \dfrac{1}{4}$$ $$p_1(X_1=2, X_2 = 2) = \dfrac{1}{4}$$

The probability distribution $p_2$ over $X_1, X_2$ is given by

$$p_2(X_1=1, X_2 = 1) = \dfrac{8}{16}$$ $$p_2(X_1=1, X_2 = 2) = \dfrac{4}{16}$$ $$p_2(X_1=2, X_2 = 1) = \dfrac{3}{16}$$ $$p_2(X_1=2, X_2 = 2) = \dfrac{1}{16}$$

Suppose $D_1, D_2$ are the datasets generated by $p_1, p_2$ respectively.

Then which dataset can I call an iid? I am guessing as $D_1$ since we can prove the random variables are independent and are identically distributed and for $D_2$, iid does not hold.

$\underline{\text{ For }D_1}$

Identical : $p_1(X_1 = x_1) = p_1(X_2 = x_2)= \dfrac{1}{2} \text{ where } x_1 = x_2 \in \{1, 2\}$
Independent: $p_1(X_1 = x_1,X_2 = x_2) = \dfrac{1}{4} = p_1(X_1 = x_1) p_1(X_2 = x_2) \text{ for } x_1, x_2 \in \{1, 2\}$

We can show that random variables $X_1, X_2$ are not iid if we consider $p_2$.

Is the iid I am discussing is different from making an iid of a dataset as answered here? If not, where am I going wrong?

  • $\begingroup$ The probability distributions used to describe i.i.d. vs non-i.i.d. are normally considered per example, whilst here it looks like you want to consider a pd over the whole dataset at once. Whilst that is possible, that is an unusual approach, and I am not quite sure what insight you hope to gain with it. Is that really what you want to ask? The pd examples you used in the linked question were clearly per example $\endgroup$ Sep 12, 2021 at 8:09
  • $\begingroup$ @NeilSlater It is difficult for me to understand what you mean by per example related probability distributions. I can image only a single pd for a whole dataset. If the random variables obey independence and identical distribution properties as $p_1$ in $D_1$, then I can recognize it as iid else non-iid as $D_2$ because of $p_2$. If possible, please recommend some material so that I can understand in detail what you are saying. $\endgroup$
    – hanugm
    Sep 12, 2021 at 8:25
  • $\begingroup$ Sorry I don't understand your comment or question. I hope someone can help answer your question. $\endgroup$ Sep 12, 2021 at 8:58

1 Answer 1


A sequence of $n$ random variables $z_{1:n} = z_1, z_2, \dots, z_n$ is i.i.d. if

  1. they are identically distributed, i.e. each random variable $z_i$ has the same distribution
  2. the joint distribution of all of them is just the product of the marginal distributions of each r.v.

So, let's imagine a thought experiment in which we throw a coin $n$ times, so you get a sequence of results, but you still don't know what those actual results are (i.e. you don't know whether $z_i$ is heads or tails), because this is just a thought experiment, so we're still talking about random variables and probability distributions and not datasets.

You might think: well, we're sampling $n$ times from the same distribution because we have only 1 coin, so we have only one r.v. In reality, you can think that each throw is associated with a different r.v. $z_i$, but that all these r.v.s, $z_1, z_2, \dots, z_n$, have the same probability distribution, for example, a Bernoulli with the same $p$ (the parameter of the Bernoulli). Now, let's say that $p = 0.5$. This means that, for a single coin toss, there's a 50% chance that the coin will be tails and a 50% chance it lands on heads. This does not have to be the case. In fact, we could also have a weird coin that prefers to land on heads, so let's say that $p = 0.7$, which is the probability it lands on heads. That's fine, and the sequence of random variables $z_{1:n}$ can still be i.i.d. How is this possible?

What's the Bernoulli pmf?

$$f_\text{marginal}(k_i;p)=p^{k_i}(1-p)^{1-k_i},$$ where $k_i\in \{0,1\}$.

So, each $z_i$ has this Bernoulli pmf, with the same $p$. This is the identically distributed part of iid.

For simplicity, let $n = 2$, so $z_{1:n} = z_{1:2} = z_1, z_2$.

So, if $z_1$ and $z_n$ are independent, we have that their joint is just the product of their marginals

\begin{align} f_\text{joint}(k_1, k_2;p) &=(p^{k_1}(1-p)^{1-k_1}) (p^{k_2}(1-p)^{1-k_2}) \\ &=p^{k_1} p^{k_2} (1-p)^{1-k_1} (1-p)^{1-k_2} \\ &=p^{k_1 + k_2} (1-p)^{2-k_1 - k_2} \\ \end{align} where $k_1 \in \{0,1\}$ is the outcome for $z_1$ and $k_2 \in \{0,1\}$ is the outcome for $z_2$.

So, as before, let's say that $p= 0.7$, then that can be written as

$$ f_\text{joint}(k_1, k_2; 0.7) = (0.7^{k_1}(1-0.7)^{1-k_1}) (0.7^{k_2}(1-0.7)^{1-k_2}) $$

If $k_1=0$ and $k_2 = 0$, we have

\begin{align} f_\text{joint}(0, 0;0.7) &= (0.7^0(1-0.7)^1) (0.7^0(1-0.7)^1) \\ &= 0.3 * 0.3 = 0.09 \end{align}

If $k_1=1$ and $k_2 = 1$, we have

\begin{align} f_\text{joint}(1, 1;0.7) &= (0.7^1(1-0.7)^0) (0.7^1(1-0.7)^0) \\ &= 0.7 * 0.7 = 0.49 \end{align}

If $k_1= 1$ and $k_2 = 0$, we have

\begin{align} f_\text{joint}(1, 1;0.7) &= (0.7^1(1-0.7)^0) (0.7^0(1-0.7)^1) \\ &= 0.7 * 0.3 = 0.21 \end{align}

If $k_1= 0$ and $k_2 = 1$, we have

\begin{align} f_\text{joint}(1, 1;0.7) &= (0.7^0(1-0.7)^1) (0.7^1(1-0.7)^0) \\ &= 0.3 * 0.7 = 0.21 \end{align}

So, the probabilities are not uniform, but we still have two independent r.v.s, because we defined their joint as the product of their marginals.

Now, set $p = 0.5$, you will see that we will have $1/4$ for all the combinations of $k_1$ and $k_2$.

Generally, the joint distribution of $n$ independent Bernoulli can be compactly written as follows

$$ f_\text{joint}(k_1, \dots, k_n; p) = p^{\sum k_i}(1-p)^{n- \sum k_i} $$

Conclusion: you cannot determine whether a sequence of r.v.s is i.i.d. by just looking at the probabilities.

  • $\begingroup$ Okay, I think I understood now. If we consider a dataset, then we should not consider features as random variables to call the dataset an iid and we should view instances of the dataset as random variables. $\endgroup$
    – hanugm
    Jan 16, 2022 at 8:05
  • 1
    $\begingroup$ @hanugm You can consider instances of the dataset as random variables, I would say. The dataset would be like the "sequence" here. $\endgroup$
    – nbro
    Jan 16, 2022 at 8:56

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