Finding a Maximum Likelihood Estimator (X=(X1,X2,...,Xn) random sample ~ E(λ))

[mihalaras]

Description
We have X=(X1,X2,...,Xn) random sample ~ E(λ)

Finding a Maximum Likelihood Estimator:


My Questions:
I have marked with red color the parts that I'm facing an issue to understand
1) I can't understand why Π {e^(-xi)} has a "/n" when we turn it into a Σ. Should it not be just "e^Σ(-xi)" ?
2) What is the logic behind finding the derivative of the first red marked term and turning it into the second red marked term? What rule do we follow to do this?
3) Basically the same as number 2, as we take the second derivative and we have the third red marked term. Again, if there is a rule that explains this, it would be very helpful.

Thank you in advance!!

 

[blamocur]

Regarding 1): I didn't find any "/n", but have "[imath]/\lambda[/imath]" comes from the definition of the exponential distribution. BTW, the previous term in that equality is incorrect, which might be the cause of your confusion.

 

[blamocur]

2) What is the logic behind finding the derivative of the first red marked term and turning it into the second red marked term? What rule do we follow to do this?

The logic of find the derivative is to help in maximizing the likelihood. Since [imath]\log[/imath] is a monotonic function its maximum is achieved for the same value of [imath]\lambda[/imath], but the expression is simpler.

 

[blamocur]

3) Basically the same as number 2, as we take the second derivative and we have the third red marked term. Again, if there is a rule that explains this, it would be very helpful.

Once you find the point with zero first derivative you use the second derivative to make sure that the critical point corresponds to the maximum (as opposed to minimum or inflection point).

P.S. Have you learned any calculus?

 

[mihalaras]

The logic of find the derivative is to help in maximizing the likelihood. Since [imath]\log[/imath] is a monotonic function its maximum is achieved for the same value of [imath]\lambda[/imath], but the expression is simpler.

Thanks for your answers. You are right about 1), however there is a slight misunderstanding on what I asked on numbers 2) and 3). I didn't explain it correctly at the beginning, so let me try to rephrase.

You explained me the theoretical reason on the steps that we follow, which I can understand.

What I struggled to understand was the practical finding of the derivative e^(Σ(-xi)/λ) which turns out to be Σxi/λ^2 . What is the formula that we follow in order to find this one out?

 

[BigBeachBanana]

Thanks for your answers. You are right about 1), however there is a slight misunderstanding on what I asked on numbers 2) and 3). I didn't explain it correctly at the beginning, so let me try to rephrase.

You explained me the theoretical reason on the steps that we follow, which I can understand.

What I struggled to understand was the practical finding of the derivative e^(Σ(-xi)/λ) which turns out to be Σxi/λ^2 . What is the formula that we follow in order to find this one out?

Is this your work or a posted solution? A few comments

• That's the incorrect density function for the exponential function (pointed out above)
• The maximum likelihood is incorrect. It should be [imath]\lambda = \dfrac{1}{\bar{x}}[/imath]
• The second derivative should be divided by [imath]\lambda^3[/imath] not [imath]\lambda^2[/imath]

 

[blamocur]

What I struggled to understand was the practical finding of the derivative e^(Σ(-xi)/λ) which turns out to be Σxi/λ^2 .

It is not the derivative of [imath]e^{\sum x_i / \lambda}[/imath] but of log of [imath]e^{\sum x_i / \lambda}[/imath]. Does it make sense now?

 

[mihalaras]

Is this your work or a posted solution? A few comments

• That's the incorrect density function for the exponential function (pointed out above)
• The maximum likelihood is incorrect. It should be [imath]\lambda = \dfrac{1}{\bar{x}}[/imath]
• The second derivative should be divided by [imath]\lambda^3[/imath] not [imath]\lambda^2[/imath]

Thank you a lot for the corrections!

It is actually from notes of a classmate of mine, so it's possible he copied something wrong.

 

[mihalaras]

It is not the derivative of [imath]e^{\sum x_i / \lambda}[/imath] but of log of [imath]e^{\sum x_i / \lambda}[/imath]. Does it make sense now

Oh yes, you are right. Thank you a lot for your time.

 

[BigBeachBanana]

Thank you a lot for the corrections!

It is actually from notes of a classmate of mine, so it's possible he copied something wrong.

See this link

 

[khansaheb]

Description
We have X=(X1,X2,...,Xn) random sample ~ E(λ)

Finding a Maximum Likelihood Estimator:

View attachment 36417
My Questions:
I have marked with red color the parts that I'm facing an issue to understand
1) I can't understand why Π {e^(-xi)} has a "/n" when we turn it into a Σ. Should it not be just "e^Σ(-xi)" ?
2) What is the logic behind finding the derivative of the first red marked term and turning it into the second red marked term? What rule do we follow to do this?
3) Basically the same as number 2, as we take the second derivative and we have the third red marked term. Again, if there is a rule that explains this, it would be very helpful.

Thank you in advance!!

It is actually from notes of a classmate of mine,

Have you found answers to the doubts/misunderstandings you had posted