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I was studying the concept of random variables when I encountered an exercise problem (from Introduction to Probability (2e) - Blitzstein & Hwang). I noticed that somebody had already asked a question regarding the exact same exercise problem here. However, it didn't address the question I had, which prompted me to ask my own.


Question:

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.


I derived solutions to both, but had a question regarding the particular solution to (b).


My Solution:

$$P(X = k\ |\ X \ge 2)\ =\ \frac{P(X = k,\ X \ge 2)}{P(X \ge 2)}$$ $$=\ \frac{P(X = k)}{P(X \ge 2)}$$ $$=\ \frac{P(X = k)}{1\ -\ (P(X = 0)\ +\ P(X = 1))}$$ $$=\ \frac{\binom{10}{k}(\frac{1}{2})^{10}}{\frac{1013}{1024}}$$ $$=\ \frac{\binom{10}{k}}{1013}$$

(If my solution is incorrect, please do let me know.)


My specific question has to do with the first line that I wrote:

How do we come to the conclusion that $P(X=k,\ X \ge 2)\ =\ P(X=k)$?

My initial thought is that this line of reasoning may come from the fact that we'd be dividing the cases to where ($k = 0,\ 1$) and ($k \ge 2$), so it wouldn't matter which particular probability we're using. Is that correct?

Any feedback is appreciated. Thank you.

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As you write $P(X=k, X \ge 2)$, I am going to assume that $k \ge 2$, otherwise the probability is $0$.

Now, if $k \ge 2$, $Pr(X =k , X \ge 2)=Pr(X =k)$ since $X=k$ have already imply that $X \ge 2$.

$$Pr(X=k, X \ge 2) = \begin{cases} 0 &, k < 2 \\ Pr(X=k) &, k \ge 2. \end{cases}$$

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  • $\begingroup$ Thanks for the feedback! :) $\endgroup$ – Seankala Oct 15 '18 at 9:51
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You are correct that the cases should be so divided.   The answer given is for $x\geq 2$, while the answer is $0$ for $x\leq 1$.

There should be an indicator for the support of the probability function.$$\mathsf P(X=k\mid X\geq 2) =\dfrac{\binom {10}k }{2^{10}-\binom{10}0-\binom{10}1}\mathbf 1_{k\geq 2, k\in\Bbb N}$$

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  • $\begingroup$ Hello, I noticed you're the answerer to the original post. I was actually wondering about the support notation in your previous answer, as well. If I'm correct, does it mean that for the condition specified (i.e. $k \ge 2$), it's 1 and otherwise 0? $\endgroup$ – Seankala Oct 15 '18 at 9:52
  • $\begingroup$ Yes, exactly, @Sean . It is an indicator function; valued $1$ when the subscripted statement is true, and $0$ otherwise. $\endgroup$ – Graham Kemp Oct 15 '18 at 13:07
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If $k\geq2$ then $\{X=k,X\geq 2\}=\{X=k\}$ so both events have the same probability:$$P(X=k,X\geq 2)=P(X=k)$$

If $k<2$ then $\{X=k,X\geq 2\}=\varnothing$ so that:$$P(X=k,X\geq 2)=P(\varnothing)=0$$

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