Continuity correction, poisson

Tarly

New Member
#1
Hi

There is something that puzzles me. I tried calculating a Poisson probability. Exactly, and with the normal distribution, using both the continuity and without the continuity correction. However the answer became much more exact if I did it not using continuity correction, do you see why?

It is a poisson distribution with parameter \(\lambda=168\).
I want to calculate \(P(X \ge 196)\).

I get that the exact answer is:
0.01876504
The answer using CLT and continuity correction :0.01693269
And without the continuity correction I get: 0.01862127

R-code:
Code:
1-ppois(195,168) #exact
1-pnorm((195.5-168)/sqrt(168)) #with correction
1-pnorm((195-168)/sqrt(168))   #without correction
Do you see why in this case it is best to not use continuity correction?
 

Dason

Ambassador to the humans
#2
Your "without correction" method isn't actually correct. If you were to go without a correction you would just look at the area greater than or equal to 196 (exactly what the probability statement says).

Code:
> 1-ppois(195,168) #exact
[1] 0.01876504
> 1-pnorm((195.5-168)/sqrt(168)) #with correction
[1] 0.01693269
> 1-pnorm((196-168)/sqrt(168)) #without correction
[1] 0.01537678
So the correction is helping.
 

Tarly

New Member
#3
Your "without correction" method isn't actually correct. If you were to go without a correction you would just look at the area greater than or equal to 196 (exactly what the probability statement says).

Code:
> 1-ppois(195,168) #exact
[1] 0.01876504
> 1-pnorm((195.5-168)/sqrt(168)) #with correction
[1] 0.01693269
> 1-pnorm((196-168)/sqrt(168)) #without correction
[1] 0.01537678
So the correction is helping.
I got 195 using this:

\(P(X \ge 196)=1-P(X <195)=1-P(\frac{X-168}{\sqrt{168}}<\frac{195-168}{\sqrt{168}})\)
\(\approx P(Z<\frac{195-168}{\sqrt{168}})\)

Did I do a mistake somewhere?
 

Dason

Ambassador to the humans
#4
You have a mistake on your first equality: It should be P(X >= 196) = 1 - P(X < 196).

But that doesn't matter - you're bringing in the discrete nature of the poisson too soon. Remember what you're doing here. You'll applying the CLT. So you're saying that you think your random variable is approximately normally distributed and you know that the mean is 168 and the variance is also 168.

So if I tell you that you have a normal random variable with a mean and variance of 168 and ask you to find P(X >= 196) what would you do? Ignore everything else - just use the fact that you have a random variable that you think is normal and you know the mean and the variance.