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.:)

  • $\begingroup$ I do not know whether arbitrary linear combination is allowed in Poisson. But when I search the net, it seems like the coefficients of the random Poisson variables have to be 1 so that the combination (which is a sum) is a Poisson random variable $\endgroup$ – Novice Oct 5 '14 at 4:47
  • $\begingroup$ I changed $Po(120)$~$N(120,120)$ to $Po(120)\sim N(120,120)$. TeX and LaTeX and MathJax were not invented by troglodytes; they can do lots of things. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 5 '14 at 4:58

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 $D-4W$.

  • $\begingroup$ Hi I will like to ask regarding the statement "W has Poisson distribution mean 1.2" Are you referring to W has Poisson distribution mean 120? $\endgroup$ – Novice Oct 5 '14 at 5:06
  • $\begingroup$ I had a "typo" there. It is $120$. $\endgroup$ – André Nicolas Oct 5 '14 at 5:07
  • $\begingroup$ Oh thank you for the clarification. We cannot times the four inside so that it is a Poisson with average mean 480? $\endgroup$ – Novice Oct 5 '14 at 5:08
  • $\begingroup$ 4W is not equal W+W+W+W? $\endgroup$ – Novice Oct 5 '14 at 5:09
  • 1
    $\begingroup$ Note that $4W$ is not Poisson. A Poisson can take on values not divisible by $4$, while $4W$ cannot. Of course $4W$ is $W+W+W+W$, but it is not a sum $W_1+W_2+W_3+W_4$ of independent Poisson random variables. However, $W$ is "approximately normal", so $4W$ is approximately normal. $\endgroup$ – André Nicolas Oct 5 '14 at 5:13

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