Let $X_1, X_2, \ldots, X_n$ be a random sample from a population with $E(X_i) = \mu$ for all $i \in \{1,\ldots, n \}$.


$ Y_i = \begin{cases} 1 & \mbox{ if } X_i < \mu \\ 0 & \mbox{ if } X_i > \mu \\ \end{cases} $

1) Determine the distribution of $Y = \sum_{i=1}^n Y_i$ (name and parameters)

Is there anyone who can give me a bit of guidance here. I think it's a binomial, but whatever $p = $ is unclear to me.

  • $\begingroup$ Let $P(Y_i = 1) = p$ and $P(Y_i = 0) = 1-p$. What's the probability that $Y_1+\ldots+Y_n = m$, i.e. that exactly $m$ out of $n$ times you got $Y_i=1$? $\endgroup$
    – fgp
    Mar 24, 2014 at 14:50
  • $\begingroup$ $Y_i$ is not entirely defined, since it is possible that $X_i=\mu$. So we need to fix the definition of $Y_i$ by saying say $Y_i=1$ if $X_i\le \mu$. No fix is needed if we are drawing from a continuous distribution. After the fix, $Y$ is binomial. We cannot say anything useful about $p$ in general. $\endgroup$ Mar 24, 2014 at 14:52
  • $\begingroup$ Duplicate of the first part of this question. $\endgroup$ Mar 25, 2014 at 13:18

2 Answers 2


Exactly, this is binomial.

Hint: Each $Y_i$ is a Bernoulli random variable with two possible results (say $1=$ success if $X>μ$ and $0$-failure if $X <μ$. The sum of $n$ independent Bernoulli with the same success probability $$p=P(X>μ)$$ is binomially distributed with parameters $n$ and $p$, in symbols $$Y \sim \mathrm{Bin} (n, p)$$ where $p=P(X>μ)$.



The $Y_i$ are iid and Bernoulli distributed. The sum of $n$ such rv's is binomially distributed.


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