Calculating variance of poisson distributed random variable












3












$begingroup$


I am calculating variance of a Poisson distributed random variable with mean $lambda$. I am doing it in the following way:



$mathbb{V}(X) = mathbb{E}(X^2) - lambda^2 \
= sum_{xgeq 0} quad x^2frac{e^{-lambda} quadlambda^x}{x!} - lambda^2\
= 0 + lambda e^{-lambda} + sum_{xgeq 2} quad x^2frac{e^{-lambda} quadlambda^x}{x!}-lambda^2\
= lambda e^{-lambda} + e^{-lambda}lambda^2sum_{xgeq 2} quad frac{lambda^{x-2}}{(x-2)!}\
=lambda e^{-lambda} + lambda^2 -lambda^2\
=lambda e^{-lambda}$



The answer is wrong. But I cannot understand what am I doing wrong in breaking the summation on line 3.



Edit: shown step 4 to answer a comment






self-study poisson-distribution arithmetic







share|cite|improve this question











$endgroup$












  • $begingroup$
    How do you get that sum over $x geq 2$ to be equal to $lambda^2$?
    $endgroup$
    – jbowman
    Feb 10 at 21:29












  • $begingroup$
    I have added a step.
    $endgroup$
    – anotherone
    Feb 10 at 21:37






  • 1




    $begingroup$
    Going from line 3 to line 4, you've assumed $x^2/x! = 1/(x - 2)!$ for $x geq 2$, which is not true. Also, your $-lambda^2$ term disappears on line 4 and then reappears on line 5.
    $endgroup$
    – Alex
    Feb 10 at 21:52








  • 5




    $begingroup$
    Often times it's easier to start by finding $E[X(X-1)]$. Also, please add the self-study tag, and I'll assist further.
    $endgroup$
    – StatsStudent
    Feb 10 at 21:55








  • 1




    $begingroup$
    @StatsStudent Yes, I found that just now. Thanks.
    $endgroup$
    – anotherone
    Feb 10 at 22:00
















3












$begingroup$


I am calculating variance of a Poisson distributed random variable with mean $lambda$. I am doing it in the following way:



$mathbb{V}(X) = mathbb{E}(X^2) - lambda^2 \
= sum_{xgeq 0} quad x^2frac{e^{-lambda} quadlambda^x}{x!} - lambda^2\
= 0 + lambda e^{-lambda} + sum_{xgeq 2} quad x^2frac{e^{-lambda} quadlambda^x}{x!}-lambda^2\
= lambda e^{-lambda} + e^{-lambda}lambda^2sum_{xgeq 2} quad frac{lambda^{x-2}}{(x-2)!}\
=lambda e^{-lambda} + lambda^2 -lambda^2\
=lambda e^{-lambda}$



The answer is wrong. But I cannot understand what am I doing wrong in breaking the summation on line 3.



Edit: shown step 4 to answer a comment






self-study poisson-distribution arithmetic







share|cite|improve this question











$endgroup$












  • $begingroup$
    How do you get that sum over $x geq 2$ to be equal to $lambda^2$?
    $endgroup$
    – jbowman
    Feb 10 at 21:29












  • $begingroup$
    I have added a step.
    $endgroup$
    – anotherone
    Feb 10 at 21:37






  • 1




    $begingroup$
    Going from line 3 to line 4, you've assumed $x^2/x! = 1/(x - 2)!$ for $x geq 2$, which is not true. Also, your $-lambda^2$ term disappears on line 4 and then reappears on line 5.
    $endgroup$
    – Alex
    Feb 10 at 21:52








  • 5




    $begingroup$
    Often times it's easier to start by finding $E[X(X-1)]$. Also, please add the self-study tag, and I'll assist further.
    $endgroup$
    – StatsStudent
    Feb 10 at 21:55








  • 1




    $begingroup$
    @StatsStudent Yes, I found that just now. Thanks.
    $endgroup$
    – anotherone
    Feb 10 at 22:00














3












3








3


1



$begingroup$


I am calculating variance of a Poisson distributed random variable with mean $lambda$. I am doing it in the following way:



$mathbb{V}(X) = mathbb{E}(X^2) - lambda^2 \
= sum_{xgeq 0} quad x^2frac{e^{-lambda} quadlambda^x}{x!} - lambda^2\
= 0 + lambda e^{-lambda} + sum_{xgeq 2} quad x^2frac{e^{-lambda} quadlambda^x}{x!}-lambda^2\
= lambda e^{-lambda} + e^{-lambda}lambda^2sum_{xgeq 2} quad frac{lambda^{x-2}}{(x-2)!}\
=lambda e^{-lambda} + lambda^2 -lambda^2\
=lambda e^{-lambda}$



The answer is wrong. But I cannot understand what am I doing wrong in breaking the summation on line 3.



Edit: shown step 4 to answer a comment






self-study poisson-distribution arithmetic







share|cite|improve this question











$endgroup$




I am calculating variance of a Poisson distributed random variable with mean $lambda$. I am doing it in the following way:



$mathbb{V}(X) = mathbb{E}(X^2) - lambda^2 \
= sum_{xgeq 0} quad x^2frac{e^{-lambda} quadlambda^x}{x!} - lambda^2\
= 0 + lambda e^{-lambda} + sum_{xgeq 2} quad x^2frac{e^{-lambda} quadlambda^x}{x!}-lambda^2\
= lambda e^{-lambda} + e^{-lambda}lambda^2sum_{xgeq 2} quad frac{lambda^{x-2}}{(x-2)!}\
=lambda e^{-lambda} + lambda^2 -lambda^2\
=lambda e^{-lambda}$



The answer is wrong. But I cannot understand what am I doing wrong in breaking the summation on line 3.



Edit: shown step 4 to answer a comment






self-study poisson-distribution arithmetic




self-study poisson-distribution arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 10 at 22:26









whuber♦

204k33443813




204k33443813










asked Feb 10 at 21:23









anotheroneanotherone

1326




1326












  • $begingroup$
    How do you get that sum over $x geq 2$ to be equal to $lambda^2$?
    $endgroup$
    – jbowman
    Feb 10 at 21:29












  • $begingroup$
    I have added a step.
    $endgroup$
    – anotherone
    Feb 10 at 21:37






  • 1




    $begingroup$
    Going from line 3 to line 4, you've assumed $x^2/x! = 1/(x - 2)!$ for $x geq 2$, which is not true. Also, your $-lambda^2$ term disappears on line 4 and then reappears on line 5.
    $endgroup$
    – Alex
    Feb 10 at 21:52








  • 5




    $begingroup$
    Often times it's easier to start by finding $E[X(X-1)]$. Also, please add the self-study tag, and I'll assist further.
    $endgroup$
    – StatsStudent
    Feb 10 at 21:55








  • 1




    $begingroup$
    @StatsStudent Yes, I found that just now. Thanks.
    $endgroup$
    – anotherone
    Feb 10 at 22:00


















  • $begingroup$
    How do you get that sum over $x geq 2$ to be equal to $lambda^2$?
    $endgroup$
    – jbowman
    Feb 10 at 21:29












  • $begingroup$
    I have added a step.
    $endgroup$
    – anotherone
    Feb 10 at 21:37






  • 1




    $begingroup$
    Going from line 3 to line 4, you've assumed $x^2/x! = 1/(x - 2)!$ for $x geq 2$, which is not true. Also, your $-lambda^2$ term disappears on line 4 and then reappears on line 5.
    $endgroup$
    – Alex
    Feb 10 at 21:52








  • 5




    $begingroup$
    Often times it's easier to start by finding $E[X(X-1)]$. Also, please add the self-study tag, and I'll assist further.
    $endgroup$
    – StatsStudent
    Feb 10 at 21:55








  • 1




    $begingroup$
    @StatsStudent Yes, I found that just now. Thanks.
    $endgroup$
    – anotherone
    Feb 10 at 22:00
















$begingroup$
How do you get that sum over $x geq 2$ to be equal to $lambda^2$?
$endgroup$
– jbowman
Feb 10 at 21:29






$begingroup$
How do you get that sum over $x geq 2$ to be equal to $lambda^2$?
$endgroup$
– jbowman
Feb 10 at 21:29














$begingroup$
I have added a step.
$endgroup$
– anotherone
Feb 10 at 21:37




$begingroup$
I have added a step.
$endgroup$
– anotherone
Feb 10 at 21:37




1




1




$begingroup$
Going from line 3 to line 4, you've assumed $x^2/x! = 1/(x - 2)!$ for $x geq 2$, which is not true. Also, your $-lambda^2$ term disappears on line 4 and then reappears on line 5.
$endgroup$
– Alex
Feb 10 at 21:52






$begingroup$
Going from line 3 to line 4, you've assumed $x^2/x! = 1/(x - 2)!$ for $x geq 2$, which is not true. Also, your $-lambda^2$ term disappears on line 4 and then reappears on line 5.
$endgroup$
– Alex
Feb 10 at 21:52






5




5




$begingroup$
Often times it's easier to start by finding $E[X(X-1)]$. Also, please add the self-study tag, and I'll assist further.
$endgroup$
– StatsStudent
Feb 10 at 21:55






$begingroup$
Often times it's easier to start by finding $E[X(X-1)]$. Also, please add the self-study tag, and I'll assist further.
$endgroup$
– StatsStudent
Feb 10 at 21:55






1




1




$begingroup$
@StatsStudent Yes, I found that just now. Thanks.
$endgroup$
– anotherone
Feb 10 at 22:00




$begingroup$
@StatsStudent Yes, I found that just now. Thanks.
$endgroup$
– anotherone
Feb 10 at 22:00










1 Answer
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$begingroup$

The Poisson distribution is one of those distributions that involves a factorial denominator. This this type of distribution, the simplest way to find the raw and central moments is to first find the expected values of the falling factorials:



$$mathbb{E}((X)_r) = mathbb{E} bigg( X(X-1) cdots (X-r+1) bigg) = sum_{x=0}^infty x(x-1) cdots (x-r+1) cdot frac{lambda^x}{x!} e^{-lambda}.$$



Have a go at solving this infinite sum (it is relatively simple) and then you will have a general form for the expected value of the falling factorials of $X$. These are all polynomials in $X$, so they can be used to find any of the raw moments or central moments by appropriate arithmetic. In your particular case you want the variance, so you will want to look at the expected values of the falling factorials up to $r=2$.






share|cite|improve this answer









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    $begingroup$

    The Poisson distribution is one of those distributions that involves a factorial denominator. This this type of distribution, the simplest way to find the raw and central moments is to first find the expected values of the falling factorials:



    $$mathbb{E}((X)_r) = mathbb{E} bigg( X(X-1) cdots (X-r+1) bigg) = sum_{x=0}^infty x(x-1) cdots (x-r+1) cdot frac{lambda^x}{x!} e^{-lambda}.$$



    Have a go at solving this infinite sum (it is relatively simple) and then you will have a general form for the expected value of the falling factorials of $X$. These are all polynomials in $X$, so they can be used to find any of the raw moments or central moments by appropriate arithmetic. In your particular case you want the variance, so you will want to look at the expected values of the falling factorials up to $r=2$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      The Poisson distribution is one of those distributions that involves a factorial denominator. This this type of distribution, the simplest way to find the raw and central moments is to first find the expected values of the falling factorials:



      $$mathbb{E}((X)_r) = mathbb{E} bigg( X(X-1) cdots (X-r+1) bigg) = sum_{x=0}^infty x(x-1) cdots (x-r+1) cdot frac{lambda^x}{x!} e^{-lambda}.$$



      Have a go at solving this infinite sum (it is relatively simple) and then you will have a general form for the expected value of the falling factorials of $X$. These are all polynomials in $X$, so they can be used to find any of the raw moments or central moments by appropriate arithmetic. In your particular case you want the variance, so you will want to look at the expected values of the falling factorials up to $r=2$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        The Poisson distribution is one of those distributions that involves a factorial denominator. This this type of distribution, the simplest way to find the raw and central moments is to first find the expected values of the falling factorials:



        $$mathbb{E}((X)_r) = mathbb{E} bigg( X(X-1) cdots (X-r+1) bigg) = sum_{x=0}^infty x(x-1) cdots (x-r+1) cdot frac{lambda^x}{x!} e^{-lambda}.$$



        Have a go at solving this infinite sum (it is relatively simple) and then you will have a general form for the expected value of the falling factorials of $X$. These are all polynomials in $X$, so they can be used to find any of the raw moments or central moments by appropriate arithmetic. In your particular case you want the variance, so you will want to look at the expected values of the falling factorials up to $r=2$.






        share|cite|improve this answer









        $endgroup$



        The Poisson distribution is one of those distributions that involves a factorial denominator. This this type of distribution, the simplest way to find the raw and central moments is to first find the expected values of the falling factorials:



        $$mathbb{E}((X)_r) = mathbb{E} bigg( X(X-1) cdots (X-r+1) bigg) = sum_{x=0}^infty x(x-1) cdots (x-r+1) cdot frac{lambda^x}{x!} e^{-lambda}.$$



        Have a go at solving this infinite sum (it is relatively simple) and then you will have a general form for the expected value of the falling factorials of $X$. These are all polynomials in $X$, so they can be used to find any of the raw moments or central moments by appropriate arithmetic. In your particular case you want the variance, so you will want to look at the expected values of the falling factorials up to $r=2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 10 at 22:52









        BenBen

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