Problem:
Leaves of a plant are examined for insects and it is
found that x_i
leaves have
precisely i insects (i=1,2,...;Sum (x_i) = N). The
number of insects
per leaf is believed
to be Poisson, except that many leaves have no
insects because they
are unsuitable for feeding and not merely because of
the chance
variation allowed for by the Poisson distribution.
The empty leaves
are therefore not counted. Show that:

(*) T=Sum_{i=2}^{infinity} (i*x_i)/N

is an unbiased estimator of the Poisson parameter
\mu, and determine
its efficiency.
(End of problem)

------------------
My try:
Let i be random variable, i=number of insects on
plant's leaf.

Does it mean that x_1 leaves have precisely 1 insect,
x_2 leaves have
precisely 2 insects, and so on...?

Put Pr=Probability.

I think that we have that Pr(i) = ( 1-exp(-\mu)
)^(-1)*exp(-\mu)*
\mu^i/i!, for i=1,2,3,...
Expectation: E_\mu(i) = \mu.

How one can calculate E_\mu(T)? What probability
distributions have N
and x_i the number of leaves which have precisely i
insects. Are x_i
and i independent?

I tried the following: (I fixed x_i and N but I do
not know whether it
is correct)
E_\mu(T) = Sum_{i=2}^{infinity} ( E_\mu(i*x_i) )/N =
(1/
N)*Sum_{i=2}^{infinity} (x_i*E_\mu(i))=
(1/N)*\mu*Sum_{i=2}^{infinity} x_i= \mu*(N-x_1)/N
which is not equal
to \mu.


If in (*) we start i from 1 instead 2 and take

(2) T_0=Sum_{i=1}^{infinity} (i*x_i)/N

which is equal to average number of insects per one
leaf,
under above assumptions (I fixed x_i and N but I do
not know whether
it is correct) we have:

E_\mu(T_0) = Sum_{i=1}^{infinity} ( E_\mu(i*x_i) )/N
= (1/
N)*Sum_{i=1}^{infinity} (x_i*E_\mu(i))=
(1/N)*\mu*Sum_{i=1}^{infinity} x_i= \mu*N/N = \mu but
it seems to be
too simple.


Does anybody have some hints how to understand this
problem and how to
solve it?

TIA

Mark

Since the counts are censored by excluding zero, we have for i>=1,

P(i) = (mu^i)*exp(-mu)/[i!*(1-exp(-mu)]

and E(xi) = N*P(i)

Summing for i>=2,

E(Sum(i*xi)/N)) = Sum(i*E(xi))/N = Sum(i*N*(u^i)*exp(-mu)/i!*(1-exp(-mu))*N) = >[exp(-mu)/(1-exp(-mu)]*Sum((mu^i)/(i-1)!) = [exp(-mu)*mu/(1-exp(-mu)]*Sum((mu^(i-1))/(i-1)!) = >[exp(-mu)*mu/(1-exp(-mu)]*[exp(mu)-1] = mu.

For the sake of brevity, I left out a few simple algebraic steps.

Jack

On Apr 5, 11:07 pm, Jack Tomsky
jtom...@ix.netcom.com> wrote:

Thanks.

Since the counts are censored by excluding zero, we
have for i>=1,

P(i) = (mu^i)*exp(-mu)/[i!*(1-exp(-mu)]

and E(xi) = N*P(i)

Could you give some precise explanation or maybe
rigorous proof that
E(xi) = N*P(i)?

Summing for i>=2,

E(Sum(i*xi)/N)) = Sum(i*E(xi))/N =
Sum(i*N*(u^i)*exp(-mu)/i!*(1-exp(-mu))*N) =
[exp(-mu)/(1-exp(-mu)]*Sum((mu^i)/(i-1)!) =
[exp(-mu)*mu/(1-exp(-mu)]*Sum((mu^(i-1))/(i-1)!) =
[exp(-mu)*mu/(1-exp(-mu)]*[exp(mu)-1] = mu.

Could you explain why E(i*xi) = i*E(xi)?

TIA

Mark


For the sake of brevity, I left out a few simple
algebraic steps.

Jack

On Apr 5, 11:07 pm, Jack Tomsky
jtom...@ix.netcom.com> wrote:

Thanks.

Since the counts are censored by excluding zero, we
have for i>=1,

P(i) = (mu^i)*exp(-mu)/[i!*(1-exp(-mu)]

and E(xi) = N*P(i)

Could you give some precise explanation or maybe
rigorous proof that
E(xi) = N*P(i)?

P(i) is the probability that a particular leaf will have i insects. Since there are N leaves, the expected number of such leaves having i insects is P(i)+P(i)+...+P(i) or N*P(i)



Summing for i>=2,

E(Sum(i*xi)/N)) = Sum(i*E(xi))/N =
Sum(i*N*(u^i)*exp(-mu)/i!*(1-exp(-mu))*N) =
[exp(-mu)/(1-exp(-mu)]*Sum((mu^i)/(i-1)!) =
[exp(-mu)*mu/(1-exp(-mu)]*Sum((mu^(i-1))/(i-1)!) =
[exp(-mu)*mu/(1-exp(-mu)]*[exp(mu)-1] = mu.

Could you explain why E(i*xi) = i*E(xi)?

i is a constant and xi is a random variable. Since i*xi is a multiple of a random variable, you can factor out the constant i. More generally, E(a+b*X) = a+b*E(X).

Jack



TIA

Mark

For the sake of brevity, I left out a few simple
algebraic steps.

Jack

On Apr 7, 5:14 pm, Jack Tomsky
jtom...@ix.netcom.com> wrote:
On Apr 5, 11:07 pm, Jack Tomsky
jtom...@ix.netcom.com> wrote:

Thanks.

Since the counts are censored by excluding
zero, we
have for i>=1,

P(i) = (mu^i)*exp(-mu)/[i!*(1-exp(-mu)]

and E(xi) = N*P(i)

Could you give some precise explanation or maybe
rigorous proof that
E(xi) = N*P(i)?

P(i) is the probability that a particular leaf will
have i insects. Since there are N leaves, the
expected number of such leaves having i insects is
P(i)+P(i)+...+P(i) or N*P(i)



Summing for i>=2,

E(Sum(i*xi)/N)) = Sum(i*E(xi))/N =
Sum(i*N*(u^i)*exp(-mu)/i!*(1-exp(-mu))*N) =
[exp(-mu)/(1-exp(-mu)]*Sum((mu^i)/(i-1)!) =
[exp(-mu)*mu/(1-exp(-mu)]*Sum((mu^(i-1))/(i-1)!)
=
[exp(-mu)*mu/(1-exp(-mu)]*[exp(mu)-1] = mu.

Could you explain why E(i*xi) = i*E(xi)?

i is a constant and xi is a random variable. Since
i*xi is a multiple of a random variable, you can
factor out the constant i. More generally, E(a+b*X)
= a+b*E(X).


Why i is a constant? In the formula
Sum_{i=2}^infinity (i*xi)/N the
letter i denotes index but
we have that P(i) = mu^i/[i! (1-exp(-mu) ] so in my
opinion xi but
also i are a random variables.
Could you please explain it?

TIA

Mark


Jack



TIA

Mark

For the sake of brevity, I left out a few
simple
algebraic steps.

Jack

P(i) is the probability that a particular leaf will
have i insects. Since there are N leaves, the
expected number of such leaves having i insects is
P(i)+P(i)+...+P(i) or N*P(i)

Summing for i>=2,

E(Sum(i*xi)/N)) = Sum(i*E(xi))/N =
Sum(i*N*(u^i)*exp(-mu)/i!*(1-exp(-mu))*N) =
[exp(-mu)/(1-exp(-mu)]*Sum((mu^i)/(i-1)!) =
[exp(-mu)*mu/(1-exp(-mu)]*Sum((mu^(i-1))/(i-1)!)
=
[exp(-mu)*mu/(1-exp(-mu)]*[exp(mu)-1] = mu.

Could you explain why E(i*xi) = i*E(xi)?

i is a constant and xi is a random variable. Since
i*xi is a multiple of a random variable, you can
factor out the constant i. More generally, E(a+b*X)
= a+b*E(X).

Why i is a constant? In the formula
Sum_{i=2}^infinity (i*xi)/N the
letter i denotes index but
we have that P(i) = mu^i/[i! (1-exp(-mu) ] so in my
opinion xi but
also i are a random variables.
Could you please explain it?

As the expected value is evaluated term by term by the sum, for each term, i is a specific index. For example, when i is specified as 10, then the only value that i can take is 10. For the next term, i is a different specific value; i.e., 11.

Jack