# Fair coin probability experiment with strange pmf

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.

## 1 Answer

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 $Y:= \sum Y_i$.

• Each Y_i can take values -1 and +1, not 0 and 1 – Adam Mar 28 '16 at 22:56
• 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 $X$. – grand_chat Mar 28 '16 at 23:00
• @Adam $\sum_{i=1}^5 X_i = -5+ 2\sum_{i=1}^5 Y_i$ so $\mathsf P(X=1) ~=~ \mathsf P(Y=(1+5)/2)$ – Graham Kemp Mar 28 '16 at 23:37