I'm confused about independence of events in the following question from Blitzstein and Hwang, chapter 3 (Random Variables):
Let X be the number of Heads in 10 fair coin tosses. Find: (a) the conditional PMF of X, given that the first two flips landed Heads, and (b) the conditional PMF of X, given that at least two flips land Heads.
There are already three answers: Prob. 24, Random Variables and their distribution - Blitzstein and Hwang and Finding the Conditional PMF of Random Variables and Probability Distribution of No. of Heads (H) when a Fair Coin is flipped 10 times (a Blitzstein and Hwang problem), but reading them hasn't helped me understand.
If I focused purely on (a), I do not see how the first two flips being heads can influence the remaining 8 flips. The way I understand it, given that the first two flips result in heads should not provide any information about further flips. That means that $$P(X=k) = \binom{8}k\frac{1}{2}^k\frac{1}{2}^{8-k}+2$$
Ie. The remaining 8 flips simply follow the binomial distribution, but we must always add +2 because we assume 2 coins have already landed heads.
My thinking extends to (b) as well: let's flip 10 coins at once, and 'at least 2 landing heads' should have no influence on the remaining 8, so the answer to (b) should be the same as the answer to (a).
I'm assuming it's a fair coin so that, for example, if I toss it 8 times and it lands heads each time, I'm not updating my prior probability ($p=\frac{1}{2}$). But if I look at this answer, and assume at least 3 coins land heads, the PMF gets bigger?
$$P(X=k|X>=3)=\frac{\binom{10}k}{968}$$
I think I'm missing or misunderstanding some fundamental piece here. I would really appreciate if someone could point out what that is, or if they could explain how they think about this problem.
Your essential reasoning for (a) is correct but your notation and subsequent expression should be modified, since $X$ is the unconditional distribution of the number of heads and is therefore binomial with parameters $n = 10$ and $p = 0.5$. So it is incorrect to say $$\Pr[X = k] = \binom{8}{k} (0.5)^k (1 - 0.5)^{8-k} + 2.$$ First of all, the support of $X$ is on the set $\{0, 1, 2, \ldots, 10\}$, so if you plug in $k = 10$, the RHS doesn't make any sense: you'd get $$\Pr[X = 10] = \binom{8}{10} (0.5)^{10} (1 - 0.5)^{-2} + 2.$$ Also, you can't just arbitrarily add $2$ to the probability mass function. Probabilities must always range between $0$ and $1$.
The conditional random variable for the total number of heads given that the first two are heads, requires us to introduce a second random variable, say $Z$, which counts the number of heads in the first two flips. $Z$ is therefore binomial with parameters $n = 2$ and $p = 0.5$, and we can then describe the desired conditional random variable as $$X \mid Z = 2$$ and the desired probability mass function is $\Pr[X = x \mid Z = 2]$. Where your reasoning is correct is that the variable $$X - 2 \mid Z = 2$$ has a binomial distribution with $n = 8$ and $p = 0.5$; i.e., $$\Pr[X - 2 = x \mid Z = 2] = \binom{8}{x} (0.5)^x (1-0.5)^{8-x}, \quad x \in \{0, 1, 2, \ldots, 8\}. \tag{1}$$ You cannot write on the LHS $\Pr[X - 2 = x]$, nor can you add $2$ to the RHS because $2$ is added to the outcome of the random variable $X$, not to the probability of observing an outcome.
We can, however, rewrite $(1)$ by noting that $$\Pr[X - 2 = x \mid Z = 2] = \Pr[X = x+2 \mid Z = 2].$$ Then letting $k = x + 2$ or equivalently $x = k-2$, we can write $$\begin{align} \Pr[X = k \mid Z = 2] &= \binom{8}{k-2} (0.5)^{k-2} (1 - 0.5)^{8-(k-2)} \\ &= \binom{10-2}{k-2} (0.5)^{k-2} (1 - 0.5)^{10-k}, \quad k \in \{2, 3, 4, \ldots, 10\}. \tag{2} \end{align}$$
This is precisely the formula given in the first link in your question.
For (b), your reasoning is incorrect. The reason is because the event on which you are conditioning in this case is that we observe "at least 2 heads," but nowhere is it stated that those two heads must occur within the first two flips. Because those two heads can occur anywhere among the $10$ flips, you cannot leverage the independence of individual flips of the coin in the way that you propose. This time, the conditional random variable of interest is $$X \mid X \ge 2,$$ that is, the total number of heads, given that at least $2$ heads were observed. The probability mass function can be calculated as follows using the definition of conditional probability:
$$\Pr[X = x \mid X \ge 2] = \frac{\Pr[(X = x) \cap (X \ge 2)]}{\Pr[X \ge 2]}. \tag{3}$$ For $x \in \{0,1\}$, the numerator of the RHS of $(3)$ is simply $0$, since $X$ cannot simultaneously equal $0$ or $1$ and be at least $2$: such an event has $0$ probability. But for any $x \ge 2$, we can see that $\Pr[(X = x) \cap (X \ge 2)] = \Pr[X = x]$, that is, whenever $x \ge 2$, the event $X = x$ is a proper subset of the event $X \ge 2$, so the desired joint probability is simply equal to the probability of the more restrictive event. Consequently, $$\Pr[X = x \mid X \ge 2] = \begin{cases} 0, & x \in \{0,1\} \\ \frac{\Pr[X = x]}{\Pr[X \ge 2]}, & x \in \{2, 3, \ldots, 10\}. \end{cases} \tag{4}$$ You already know how to express $\Pr[X = x]$ and $\Pr[X \ge 2]$ from the binomial probability mass function, so the rest is simply computation.