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I found the following question in a past exam paper and I would like to ask how to solve it as I can't find anything in the notes related to it:

If three samples taken from Exponential(lambda) are 0.1, 0.5 
and 0.9, what's the MLE for lambda?

I don't really understand how I'm supposed to deduct it from such little information?

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let me know if anything in my answer below was not clear. Hope that helps! –  Justin Aug 6 '12 at 0:37
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2 Answers

up vote 1 down vote accepted

Consider the definition of the likelihood function for a statistical model. Here, $\theta = \lambda ,$ the unknown parameter of the distribution in question. Assuming your samples $X_1 = 0.1, X_2 = 0.5, X_3 =0.9,$ are independent, we have that the likelihood function is $f_{\lambda } (X_1, X_2, X_3) = \lambda^3 e^{-\lambda (X_1 + X_2 + X_3)}.$ Taking logarithms gives the log-likelihood function of the data; $\mathcal{L}_3 (\lambda ) = 3\log \lambda - \lambda 3\overline{x},$ where $\overline{x} = 0.5$ is the sample mean. The function is maximized at $\hat{\lambda } = \frac{1}{\overline{x}} = 2.$

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Could you please explain why the log-likelihood function is the one you mentioned? And is the value of lambda each of the values I mentioned in my post? –  Sorin Cioban Aug 5 '12 at 21:22
    
@Sorin Take a look at my revised post. I think this will clarify some aspects of the argument. –  user17794 Aug 5 '12 at 22:33
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Let's start with what we know, which is that the probability distribution function (pdf) for this problem is:

$f_X(x)=\lambda e^{-\lambda x}$ for $x\ge 0$ (and 0 otherwise)

To obtain the joint density function (since the observations are independent), we simply take the product of the individual pdfs:

$f(x_1,x_2,...,x_n)=\prod_{i=1}^n f(x_i)=\prod_{i=1}^n \lambda e^{-\lambda x_i}$

(in your example, we have 3 "x's" and so the joint pdf is:)

$f(.1,.5,.9)=\lambda^3e^{-1.5\lambda}$

Then, take the log of this function:

$=n\ln\lambda-1.5\lambda$

Now differentiate with respect to lambda:

$\frac{\partial}{\partial\lambda}\ln L(\lambda)=\frac{n}{\lambda}-1.5$

and solve for lambda (which we will now denote $\hat{\lambda}$ because it is an estimator (specifically the MLE).

Here you get $\hat{\lambda}=2$

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