I have a question regarding sum of Poisson random variables and using normal approximation:
The number of whales spotted per day has Poisson Distribution $Po(1.2)$.
The number of dolphins spotted per day has Poisson Distribution $Po(4.5)$.
The two distributions are independent.
For 100 consecutive days in a year, use approximation to find the probability that at least four times as many dolphins as whales are spotted in a year.
Let D represents the number of Dolphins, W represents the number of Whales.
I know that sum of independent Poisson random variables is still Poisson.
If suppose I am considering the random variable $X=4W-D$. (But I am not sure whether minus is allowed here)
Then X follows Poisson Distribution $Po(30)$ which the $30$ is due to $100(1.2(4)-4.5)$.
Then I am finding $P(X\leq0)$, which $X$ will have mean and variance $30$. Then I use normal approximation by doing the necessary continuity correction and so on.
However, if I do normal approximation directly first on the respective Poisson Distribution:
$Po(120)\sim N(120,120)$, $N$ is the normal distribution with mean $120$, variance $120$.
$Po(450)\sim N(450,450)$, $N$ is the normal distribution with mean $120$, variance $120$.
Defining $X=4W-D$.
Therefore $X$ follows $N(30,2370)$ and I am finding $P(X\leq0)$ using continuity correction and so on.
Both approaches yield very different answers.
Can someone rectify the error or explain the difference?
Thank you.:)
Best Answer
Let $D$ be the number of dolphins, and $W$ the number of whales. Let $$Y=D-4W.$$ We want to approximate the probability that $Y\ge 0$. Note that $D$ has Poisson distribution mean (and variance) $450$, while $W$ has Poisson distribution mean and variance $120$. The parameters are large, so we can moderately safely take $D$ and $W$ to have nearly normal distribution. Then by independence $Y$ has roughly normal distribution, mean $E(D)-4E(W)$, and variance $\text{Var}(D)+16\text{Var}(W)$.
I assume you can now finish.
Remark: The difference of Poisson random variables is nowhere near Poisson. For one thing, a random variable with Poisson distribution can never be negative, while a difference of Poisson random variables certainly can be negative.
Note that your $X$ is not what we want to look at, we want $D-4W$.