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The lifetime of a type of component has an exponential distribution with rate λ per hour. Ten of these components were tested but the only recorded results were that 3 components had failed within 100 hours and 7 had survived that time.

I have been asked to

(a) Find the maximum likelihood estimate of λ.

(b) Find the approximate standard error of this estimator.

Can you help me please?

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2 Answers 2

The probability of a component's surviving $100$ hours is $e^{-100\lambda}$. So the probability of the observed outcome, given $\lambda$ is $(e^{-100\lambda})^7 (1-e^{-100\lambda})^3$. So the likelihood function is $$ L(\lambda) = (e^{-100\lambda})^7 (1-e^{-100\lambda})^3 $$ and its logarithm is $$ \ell(\lambda) = -700\lambda + 3\log(1-e^{-100\lambda}). $$ Differentiation yields $$ -700 + \frac{3(100e^{-100\lambda})}{1-e^{-100\lambda}}. $$ That is $0$ if its product with $1-e^{-100\lambda}$ is $0$, thus if $$ -700(1-e^{-100\lambda}) + 300e^{-100\lambda}=0. $$ So $$ 1000e^{-100\lambda} = 700. $$ Can you get the answer from there?

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Hi, thanks for your reply, can you just clarify how you got the initial probability function?? –  steve Sep 28 '12 at 3:36
    
The term "exponential distribution" might be construed to mean that the density function is exponential, i.e. $f(x) = \text{constant} \cdot a^x$ for $x>0$. That's the same as $\text{constant}\cdot e^{\text{constant}\cdot x}$, where the "constant" in the exponent is $-\log_e a=-\lambda x$. If one computes the expected value $\int_0^\infty xf(x)\,dx$, one gets $1/\lambda$, so the expected waiting time until the first awaited occurrence is $1/\lambda$ and the rate is $\lambda$ per unit time. The probability of exceeding $x$ units of time is then $\int_0^x f(u)\,du=e^{-\lambda x}$. –  Michael Hardy Sep 28 '12 at 5:20

Another way of looking at the problem is that we have observed $7$ occurrences of an event of unknown probability $p$ on $10$ independent trials of the experiment. The maximum-likelihood estimate $\hat{p}$ of the unknown quantity $p$ is thus the observed relative frequency, that is, $$\hat{p} = \frac{7}{10}.$$ In this case, $p$ is of the form $\exp(-1000\lambda)$ and thus the maximum-likelihood estimate of $\lambda$ is the solution to the equation $$\exp(-100\hat{\lambda}) = \frac{7}{10}.$$

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