0
$\begingroup$

I have a question regarding an experiment where 5 fair coins are flipped, but the random variable has a quirk and is throwing me off. Fair in this case means the probability of success is p = 0.5

Each coin is labeled with +1 on the heads side and -1 on the tails side. If the random variable X is the sum of the outward facing labels after each coin is flipped, then X({HHTTH}) = 1+1-1-1+1 = 1`.

What is the probability mass function of the random variable X?

I need the pmf in order to calculate the mean, variance, and standard deviation, but I've become used to X being the number of successes, where tails is labeled 0 but in this problem, tails is marked -1.

My attempt at solving this would be to represent each coin as an independent Bernoulli trial, leading to the number of successes following a binomial distribution, but this doesn't accurately represent the random variable. Any help would be immensely appreciated.

$\endgroup$
2
$\begingroup$

Do you need to compute the pmf? If $Y_1, Y_2,\ldots, Y_5$ are the outcomes of the Bernoulli trials you've described (so each $Y_i$ takes values $0$ and $1$), then $X_i = 2Y_i-1$ are the variables you're observing. You want the mean and variance of $X:=\sum X_i$. Hint: You should be able to deduce this from the mean and variance of $Y:= \sum Y_i$.

$\endgroup$
  • $\begingroup$ Each Y_i can take values -1 and +1, not 0 and 1 $\endgroup$ – Adam Mar 28 '16 at 22:56
  • 2
    $\begingroup$ I'm using $X_i$ to represent the $\pm1$ variables. The $Y_i$ are new variables, a set of Bernoulli trials taking values $0$ and $1$. The $Y$'s will help you get to $X$. $\endgroup$ – grand_chat Mar 28 '16 at 23:00
  • $\begingroup$ @Adam $\sum_{i=1}^5 X_i = -5+ 2\sum_{i=1}^5 Y_i $ so $\mathsf P(X=1) ~=~ \mathsf P(Y=(1+5)/2)$ $\endgroup$ – Graham Kemp Mar 28 '16 at 23:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.