Approximating Poisson as Normal

If I had a Poisson distribution $X \sim \text{Po}(\lambda)$ and I wish to approximate it with $x \sim N(\lambda, \lambda)$.

Say I want to find $\mathbb{P}(X > x)$ for some non-natural $x$, e.g: $\frac{250}{3} \approx 83.3$.

Assuming that $x > \lambda$, would the continuity correction be ?

In the case of $x = \frac{250}{3} \approx 83.3$, I understand that it would be .

What if it was $x = 83.6$, would it be the same as the above or would the continuity correction be $\displaystyle \mathbb{P}\left(X > \frac{84.5 - \lambda}{\sqrt{\lambda}}\right)$ or $\displaystyle \mathbb{P}\left(\frac{84 - \lambda}{\sqrt{\lambda}}\right)$?

I'm not entirely sure how to visualise this - do I draw the usual rectangle centred around 83 and then continuity correct from there, in which case x=83.6 would yield a continuity correction of x' = 83.5?

I claim that X can only take integer values, so round up, e.g. X>83.3 actually means X >= 84. Now apply the continuity correction in the usual way, so the bar for 84 should be included, so the normal distribution gives x >= 83.5 (or x>83.5, it doesn't matter which since x is continuous).

If I had a Poisson distribution $X \sim \text{Po}(\lambda)$ and I wish to approximate it with $x \sim N(\lambda, \lambda)$.

Say I want to find $\mathbb{P}(X > x)$ for some non-natural $x$, e.g: $\frac{250}{3} \approx 83.3$.

Assuming that $x > \lambda$, would the continuity correction be ?

In the case of $x = \frac{250}{3} \approx 83.3$, I understand that it would be .

What if it was $x = 83.6$, would it be the same as the above or would the continuity correction be $\displaystyle \mathbb{P}\left(X > \frac{84.5 - \lambda}{\sqrt{\lambda}}\right)$ or $\displaystyle \mathbb{P}\left(\frac{84 - \lambda}{\sqrt{\lambda}}\right)$?

I'm not entirely sure how to visualise this - do I draw the usual rectangle centred around 83 and then continuity correct from there, in which case x=83.6 would yield a continuity correction of x' = 83.5?

Okay, that makes a ton of sense. Just to check to see if I understand:

Let's say I had $x < \lambda$ and x = 25.5, and I want P(X < 25.5), then that's the same thing as P(X <= 25) (round down) so continuity correct to X < 24.5 because I need to include the bar.

Is that good?

Okay, that makes a ton of sense. Just to check to see if I understand:

Let's say I had $x < \lambda$ and x = 25.5, and I want P(X < 25.5), then that's the same thing as P(X <= 25) (round down) so continuity correct to X < 24.5 because I need to include the bar.

Is that good?

If I had a Poisson distribution $X \sim \text{Po}(\lambda)$ and I wish to approximate it with $x \sim N(\lambda, \lambda)$.

Say I want to find $\mathbb{P}(X > x)$ for some non-natural $x$, e.g: $\frac{250}{3} \approx 83.3$.

Assuming that $x > \lambda$, would the continuity correction be ?

In the case of $x = \frac{250}{3} \approx 83.3$, I understand that it would be .

What if it was $x = 83.6$, would it be the same as the above or would the continuity correction be $\displaystyle \mathbb{P}\left(X > \frac{84.5 - \lambda}{\sqrt{\lambda}}\right)$ or $\displaystyle \mathbb{P}\left(\frac{84 - \lambda}{\sqrt{\lambda}}\right)$?

I'm not entirely sure how to visualise this - do I draw the usual rectangle centred around 83 and then continuity correct from there, in which case x=83.6 would yield a continuity correction of x' = 83.5?

Ok until the last move. You haven't included the bar. X<=25 becomes X<25.5 when you change to the normal distribution.

I think it would be clearer to use a different letter when you change to the normal distribution, e.g. X ~ Po(L), Y ~ N(L,L)

not sure how you got x = 250/3 earlier... poisson is all about integers

Latex skillz.

But yeah, integers. Those bad boys.

in a Poisson you cannot have anything but x=0,1,2,3,...

You talk too much. Have I mentioned that before?

'Pot kettle black' springs to mind. Now stop messaging me and go revise.

Are you sure? I'm saying that x is less than the mean - so including the bar means moving close to the mean, i.e: rightwards on the 'x'-axis.



Okay, thanks. I'll keep that in mind. I'd been meaning to ask about that - is there notation for saying that X is approximately distributed by Y? Is $X \approx Y$ standard? It looks out of place, so I'm going to guess no.

Are you sure? I'm saying that x is less than the mean - so including the bar means moving close to the mean, i.e: rightwards on the 'x'-axis.
Okay, thanks. I'll keep that in mind. I'd been meaning to ask about that - is there notation for saying that X is approximately distributed by Y? Is $X \approx Y$ standard? It looks out of place, so I'm going to guess no.

Whether or not you include the bar depends on the type of inequality sign, not on which side of the mean you are.

See if the example in the attachment helps.

I don't know how number lines work, apparently. Stupid mistake, thanks for clearing up the thing though!

We need a "like" button.