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Approximating Poisson as Normal Watch

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    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 \displaystyle \mathbb{P}\left(X > \frac{\floor{x} + \frac{1}{2} - \lambda}{\sqrt{\lambda}}\right)?

    In the case of x = \frac{250}{3} \approx 83.3, I understand that it would be \displaystyle \matbb{P}\left(\frac{83.5 - \lambda}{\sqrt{\lambda}}\right).

    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?
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    (Original post by Zacken)
    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 \displaystyle \mathbb{P}\left(X > \frac{\floor{x} + \frac{1}{2} - \lambda}{\sqrt{\lambda}}\right)?

    In the case of x = \frac{250}{3} \approx 83.3, I understand that it would be \displaystyle \matbb{P}\left(\frac{83.5 - \lambda}{\sqrt{\lambda}}\right).

    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).
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    (Original post by tiny hobbit)
    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).
    Okay, that makes a ton of sense. Just to check to see if I understand:

    Let's say I had x &lt; \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?
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    (Original post by Zacken)
    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 &gt; x) for some non-natural x, e.g: \frac{250}{3} \approx 83.3.

    Assuming that x &gt; \lambda, would the continuity correction be \displaystyle \mathbb{P}\left(X &gt; \frac{\floor{x} + \frac{1}{2} - \lambda}{\sqrt{\lambda}}\right)?

    In the case of x = \frac{250}{3} \approx 83.3, I understand that it would be \displaystyle \matbb{P}\left(\frac{83.5 - \lambda}{\sqrt{\lambda}}\right).

    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 &gt; \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?
    in a Poisson you cannot have anything but x=0,1,2,3,...
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    (Original post by Zacken)
    Okay, that makes a ton of sense. Just to check to see if I understand:

    Let's say I had x &lt; \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?
    :borat:

    not sure how you got x = 250/3 earlier... poisson is all about integers
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    (Original post by Zacken)
    Okay, that makes a ton of sense. Just to check to see if I understand:

    Let's say I had x &lt; \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?
    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)
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    (Original post by Zacken)
    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 &gt; x) for some non-natural x, e.g: \frac{250}{3} \approx 83.3.

    Assuming that x &gt; \lambda, would the continuity correction be \displaystyle \mathbb{P}\left(X &gt; \frac{\floor{x} + \frac{1}{2} - \lambda}{\sqrt{\lambda}}\right)?

    In the case of x = \frac{250}{3} \approx 83.3, I understand that it would be \displaystyle \matbb{P}\left(\frac{83.5 - \lambda}{\sqrt{\lambda}}\right).

    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 &gt; \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?
    Latex skillz.

    But yeah, integers. Those bad boys.
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    (Original post by tiny hobbit)
    Ok until the last move. You haven't included the bar. X<=25 becomes X<25.5 when you change to the normal distribution.
    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.

    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)
    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.
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    (Original post by the bear)
    :borat:

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

    It was part of a question where the scenario was something like 1 item is m pounds.

    Find the probability that you earn 250/3 * m pounds for some number m I can't remember. In hindsight, it was fairly obvious that I needed to round up. Poisson. :facepalm:

    Thanks!
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    (Original post by C0pper)
    Latex skillz.

    But yeah, integers. Those bad boys.
    You talk too much. Have I mentioned that before?
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    (Original post by TeeEm)
    in a Poisson you cannot have anything but x=0,1,2,3,...
    :yep: thanks!
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    (Original post by Zacken)
    You talk too much. Have I mentioned that before?
    'Pot kettle black' springs to mind. Now stop messaging me and go revise.
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    (Original post by C0pper)
    'Pot kettle black' springs to mind. Now stop messaging me and go revise.
    No.
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    (Original post by Zacken)
    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.
    Easiest way to think of continuity corrections is just to imagine them as you would with ranges for GCSE. For example, if you had that x>=25, then if you took the case that x =25 to 2 significant figures than the true value of x is 24.5<=x<25.5. Therefore if x>=24.5 in continuous , x>= 25 in poisson so you should easily be able to see which way the correction should go
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    (Original post by Zacken)
    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.
    Attached Files
  1. File Type: docx III Normal distribution as an approximation to the Poisson distribution.docx (190.5 KB, 37 views)
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    (Original post by tiny hobbit)
    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.
    :facepalm: I don't know how number lines work, apparently. Stupid mistake, thanks for clearing up the thing though!
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    (Original post by Zacken)
    :facepalm: I don't know how number lines work, apparently. Stupid mistake, thanks for clearing up the thing though!
    We need a "like" button.
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    (Original post by tiny hobbit)
    We need a "like" button.
    :lol:

    Thanks again.
 
 
 
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