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How do we obtain an estimate for $textbfP(Xgeq 1)$ where $XsimtextExp(lambda)$?

2

$begingroup$

Consider a simple sample $X_1,X_2,ldots,X_n$ whose distribution is given by $Xsim Exp(lambda)$.

(a) Determine an estimator for $lambda$ according to the method of moments.

(b) Determine another estimator for $lambda$ different from the previous one.

(c) Determine an estimate of $textbfP(Xgeq 1)$ in accordance to the method of moments.

MY ATTEMPT

As to the first case, we have
beginalign*
frac1lambda = textbfE(X) Rightarrow hatlambda = fracndisplaystylesum_k=1^nX_k
endalign*

As to the second case, we have
beginalign*
frac1lambda^2 = textbfVar(X) = textbfE(X^2) - textbfE(X)^2 Rightarrow hatlambda = left[frac1nsum_k=1^nX^2_k - left(frac1nsum_k=1^nX_kright)^2right]^-1/2
endalign*

EDIT

In the third case, we have
beginalign*
textbfP(Xgeq 1) = 1 - textbfP(X < 1) = 1 - exp(-1timeslambda) = 1 - exp(-lambda)
endalign*

Now it suffices to substitute $lambda$ for any of the above expressions.

Could someone double-check my reasoning? Thanks in advance!

statistics exponential-distribution

share|cite|improve this question

edited 2 days ago

user1337

asked Apr 8 at 1:11

user1337user1337

48210

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Exponential distribution is continuous, but you're treating it as discrete.... is there a typo?
  $endgroup$
  – Lee David Chung Lin
  Apr 8 at 2:00
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed, you are right. I was thinking about the poisson distribution when I wrote it. Thanks for the observation. I will edit it.
  $endgroup$
  – user1337
  Apr 8 at 2:08
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What have you tried? You know, don't you, that this is not a site for us to do your homework for you?
  $endgroup$
  – David G. Stork
  Apr 8 at 2:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Plug the estimator in $(a)$ into $e^-lambda$.
  $endgroup$
  – d.k.o.
  Apr 8 at 2:27
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, David. I know it. I just forgot to mention my attempts.
  $endgroup$
  – user1337
  Apr 8 at 2:38
  

  
  
  
  

 | 
show 6 more comments

2

$begingroup$

Consider a simple sample $X_1,X_2,ldots,X_n$ whose distribution is given by $Xsim Exp(lambda)$.

(a) Determine an estimator for $lambda$ according to the method of moments.

(b) Determine another estimator for $lambda$ different from the previous one.

(c) Determine an estimate of $textbfP(Xgeq 1)$ in accordance to the method of moments.

MY ATTEMPT

As to the first case, we have
beginalign*
frac1lambda = textbfE(X) Rightarrow hatlambda = fracndisplaystylesum_k=1^nX_k
endalign*

As to the second case, we have
beginalign*
frac1lambda^2 = textbfVar(X) = textbfE(X^2) - textbfE(X)^2 Rightarrow hatlambda = left[frac1nsum_k=1^nX^2_k - left(frac1nsum_k=1^nX_kright)^2right]^-1/2
endalign*

EDIT

In the third case, we have
beginalign*
textbfP(Xgeq 1) = 1 - textbfP(X < 1) = 1 - exp(-1timeslambda) = 1 - exp(-lambda)
endalign*

Now it suffices to substitute $lambda$ for any of the above expressions.

Could someone double-check my reasoning? Thanks in advance!

statistics exponential-distribution

share|cite|improve this question

edited 2 days ago

user1337

asked Apr 8 at 1:11

user1337user1337

48210

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Exponential distribution is continuous, but you're treating it as discrete.... is there a typo?
  $endgroup$
  – Lee David Chung Lin
  Apr 8 at 2:00
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed, you are right. I was thinking about the poisson distribution when I wrote it. Thanks for the observation. I will edit it.
  $endgroup$
  – user1337
  Apr 8 at 2:08
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What have you tried? You know, don't you, that this is not a site for us to do your homework for you?
  $endgroup$
  – David G. Stork
  Apr 8 at 2:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Plug the estimator in $(a)$ into $e^-lambda$.
  $endgroup$
  – d.k.o.
  Apr 8 at 2:27
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, David. I know it. I just forgot to mention my attempts.
  $endgroup$
  – user1337
  Apr 8 at 2:38
  

  
  
  
  

 | 
show 6 more comments

$begingroup$

Consider a simple sample $X_1,X_2,ldots,X_n$ whose distribution is given by $Xsim Exp(lambda)$.

(a) Determine an estimator for $lambda$ according to the method of moments.

(b) Determine another estimator for $lambda$ different from the previous one.

(c) Determine an estimate of $textbfP(Xgeq 1)$ in accordance to the method of moments.

MY ATTEMPT

As to the first case, we have
beginalign*
frac1lambda = textbfE(X) Rightarrow hatlambda = fracndisplaystylesum_k=1^nX_k
endalign*

As to the second case, we have
beginalign*
frac1lambda^2 = textbfVar(X) = textbfE(X^2) - textbfE(X)^2 Rightarrow hatlambda = left[frac1nsum_k=1^nX^2_k - left(frac1nsum_k=1^nX_kright)^2right]^-1/2
endalign*

EDIT

In the third case, we have
beginalign*
textbfP(Xgeq 1) = 1 - textbfP(X < 1) = 1 - exp(-1timeslambda) = 1 - exp(-lambda)
endalign*

Now it suffices to substitute $lambda$ for any of the above expressions.

Could someone double-check my reasoning? Thanks in advance!

statistics exponential-distribution

share|cite|improve this question

edited 2 days ago

user1337

asked Apr 8 at 1:11

user1337user1337

48210

$endgroup$

share|cite|improve this question

edited 2 days ago

user1337

asked Apr 8 at 1:11

user1337user1337

48210

share|cite|improve this question

edited 2 days ago

user1337

asked Apr 8 at 1:11

user1337user1337

48210

asked Apr 8 at 1:11

user1337user1337

48210

asked Apr 8 at 1:11

user1337user1337

48210