Determine a distribution with no parameters?

504 Views Asked by At

I'm confused by this question and I was hoping for some guidance some one to point me in the right direction

Let $X_1.........X_n$ be a random sample from a population with mean $\mu$, that is $E(X_i)=\mu$ for all $1 \le i \le n$. Define

$$Y_i = 1 \ if \ X_i \lt \mu \ and \ Y_i =0 \ otherwise$$

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

b)Determine the mean and variance of $Y$

c)Compute $P(Y \le 30)$ for special case of $n=49$ and $X_i..........X_n$ $iid$ $Exp \sim (\lambda=3)$

1

There are 1 best solutions below

3
On BEST ANSWER

For the last problem, if $X_i$ has parameter $\lambda$, then the mean $\mu$ is $\frac{1}{\lambda}$. Thus $$\Pr(X_i\lt \mu)=\int_0^{1/\lambda}\lambda e^{-\lambda x}\,dx=1-e^{-1}.$$

Our random variable $Y$ therefore has binomial distribution, $n=49$, $p=1-e^{-1}$. The probability that $Y\le 30$ is equal to $$\sum_{k=0}^{30} \binom{49}{k}(1-e^{-1})^k (e^{-1})^{49-k}.$$ This sum can nowadays be easily computed by software, including Wolfram Alpha. We can also approximate it. The distribution of $Y$ is not too far from normal, mean $49(1-e^{-1})$, variance $49(1-e^{-1})(e^{-1})$.

Remark: In the general case, $Y$ is the sum of $n$ independent identically distributed Bernoulli random variables. So it has binomial distribution, with parameters $n$ and $p$, where $p=\Pr(X_i\le \mu)$. Without further information about the distribution of the $X_i$, we cannot determine $p$.

More informally, if $X_i\lt \mu$, call that a success. Then $Y$ is the number of successes in $n$ independent repetitions of an experiment, so it has binomial distribution.