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$X $ follows $ Po(80)$.

I used a normal approximation to get $P(55\leq X\leq 75)$ which I got correct.

I need to find $P(X=80)$. I tried the Poisson distribution directly; however, my calculator shows an error when calculating $80^{80}$.

The correct answer is 0.0446.

Do we need to use normal distribution? Thanks.

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  • $\begingroup$ I've added the self-study tag because your phrasing pretty clearly indicates it falls under self-study, but you should feel free to remove it (or to object and I will remove it). $\endgroup$ – Glen_b -Reinstate Monica Apr 28 '14 at 4:20
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Here I discuss several approaches. If I wasn't near my computer (on which I'd just do dpois(80,80) in R), I'd probably use the first one below.

1) You can help your calculator out a bit by doing some algebra first, and trying Stirling's approximation for $80!$

You can try taking the series for the Stirling approximation out several terms but it won't be necessary in this case, just use the "ordinary" approximation ($n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$), which hugely simplifies the calculation, and has about 3 significant figure accuracy in this region (which will be sufficient). [Taking one more term in the series - $n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n(1+\frac{1}{12n})$ - gives about six figure accuracy.]

So take the formula for the probability:

$$P(X=80) = \frac{\exp(-80) 80^{80}}{80!}$$

Then just replace 80! by the above approximation, and cancel out all those large terms, leaving a very simple calculation. (I bet you can do it from there.)

2) The "right" way to do that is to use a function (or a series, if you are working on a calculator) that returns a log-factorial (or log-gamma-function), and work on the log-scale until the final step ... which gives the "exact" answer of about 0.04455667. While quite accurate, this may involve a bit of work unless you're using a computer.

3) If you use the normal approximation, the usual way would be to use the continuity correction. Let $X$ be the original Poisson, and let $X^*$ be an approximating normal ($N(80,(\sqrt{80})^2$), and let $Z$ be a standard normal.

$$P(X=80)\approx P(79.5<X^*<80.5)$$ $$= P(\frac{79.5-80}{\sqrt{80}}<\frac{X^*-80}{\sqrt{80}}<\frac{80.5-80}{\sqrt{80}})$$ $$= P(\frac{-0.5}{\sqrt{80}}<Z<\frac{0.5}{\sqrt{80}})$$ $$= P(-0.055902<Z<0.055902)$$ $$= 2\,P(0<Z<0.055902)$$ $$= 2\times 0.02229$$ $$= 0.04458$$

... which is also sufficiently accurate.

4) Another way to do the approximation would be to approximate it by $f_{X^*}(80)\times 1\approx 0.0446$, but you have to be careful to justify that properly (refer to the diagrams and discussion at the link on continuity corrections I gave above).

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    $\begingroup$ +1 Even Excel has a log gamma function (GAMMALN) for such purposes, suggesting how useful (and basic) taking logarithms is. I'm sure GAMMALN is implemented by means of the asymptotic expansion of $\Gamma$ (generalizing Stirling's approximation). Concerning (4), it might be a little clearer to write your approximation as $f_{X^{*}}(80) \times 1$ or even, quite suggestively, $\int_{80-1/2}^{80+1/2}\left(f_{X^{*}}(80) + O(x-80)^2\right) dx.$ $\endgroup$ – whuber Apr 23 '14 at 14:44
  • $\begingroup$ @whuber I even typed $f_{X^*}(80)\times 1$ at one point but deleted it because I was worried it was perhaps confusing. I think you're right that it is clearer, since it suggests the sort of justification that I was talking about. The integral at the end there explains the whole thing rather neatly, though. $\endgroup$ – Glen_b -Reinstate Monica Apr 23 '14 at 14:52
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You could get a better calculator or use logarithms for large powers: $$\log_{10}\left(80^{80}\right)=80\times \log_{10}\left(80\right)\approx 152.247199$$ $$80^{80} \approx 10^{0.247199} \times 10^{152} \approx 1.76685\times 10^{152}$$

If you used a normal approximation, you have two choices:

  • the density at $x=80$ of a normal distribution with mean $80$ and standard deviation $\sqrt{80}$, which is $\tfrac{1}{\sqrt{160 \pi}}$ or about $0.044603$
  • the probability of a value between $79.5$ and $80.5$ from a normal distribution with mean $80$ and standard deviation $\sqrt{80}$, which is about $0.044580$

The actual value from a Poisson distribution is closer to $0.044557$ so these approximations are not bad.

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$\ln(Poisson_{80}(80))=\exp(\ln(Poisson_{80}(80)))=\exp(80 \ln 80-80-\ln(80!))$

For $\ln(80!)$ use Ramanujan's approximation.

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