Chapter 2: Problem 22
If a fair coin is successively flipped, find the probability that a head first appears on the fifth trial.
Over 30 million students worldwide already upgrade their learning with Vaia!
All the tools & learning materials you need for study success - in one app.
Get started for free
Let \(X_{1}, X_{2}, \ldots, X_{10}\) be independent Poisson random variables with mean \(1 .\) (a) Use the Markov inequality to get a bound on \(P\left\\{X_{1}+\cdots+X_{10} \geq 15\right\\}\). (b) Use the central limit theorem to approximate \(P\left(X_{1}+\cdots+X_{10} \geq 15\right\\}\).
Let \(X_{1}, X_{2}, X_{3}\), and \(X_{4}\) be independent continuous random
variables with a common distribution function \(F\) and let
$$
p=P\left\\{X_{1}
If \(X\) is a nonnegative integer valued random variable, show that (a) $$ E[X]=\sum_{n=1}^{\infty} P[X \geq n\\}=\sum_{n=0}^{\infty} P(X>n\\} $$ Hint: Define the sequence of random variables \(I_{n}, n \geq 1\), by $$ I_{n}=\left\\{\begin{array}{ll} 1, & \text { if } n \leq X \\ 0, & \text { if } n>X \end{array}\right. $$ Now express \(X\) in terms of the \(I_{n}\). (b) If \(X\) and \(Y\) are both nonnegative integer valued random variables, show that $$ E[X Y]=\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} P(X \geq n, Y \geq m) $$
If the coin is assumed fair, then, for \(n=2\), what are the probabilities associated with the values that \(X\) can take on?
Suppose that we want to generate a random variable \(X\) that is equally likely to be either 0 or 1 , and that all we have at our disposal is a biased coin that, when flipped, lands on heads with some (unknown) probability \(p\). Consider the following procedure: 1\. Flip the coin, and let \(0_{1}\), either heads or tails, be the result. 2\. Flip the coin again, and let \(0_{2}\) be the result. 3\. If \(0_{1}\) and \(0_{2}\) are the same, return to step 1 . 4\. If \(0_{2}\) is heads, set \(X=0\), otherwise set \(X=1\). (a) Show that the random variable \(X\) generated by this procedure is equally likely to be either 0 or 1 . (b) Could we use a simpler procedure that continues to flip the coin until the last two flips are different, and then sets \(X=0\) if the final flip is a head, and sets \(X=1\) if it is a tail?
What do you think about this solution?
We value your feedback to improve our textbook solutions.
