I cannot seem to get he same answer as the book - book answer is: 0.122
Use the sample, (8.2, 9.1, 10.6, 4.9) to calculate the maximum likelihood estimate for $\lambda$ in the exponential pdf.
$$f_Y(y;\lambda)= \lambda e^{-\lambda y} \;, \quad y \ge 0$$
$$L(\lambda) = \prod_{i=1}^{n}\lambda e^{-\lambda y}$$
$$=\lambda^n e^{-\lambda \sum_{i=1}^{n}y_i}$$
$$\text{Let } T = \frac{d\big[ln(L(\lambda))\big]}{d\lambda} = \frac{n}{\lambda} - \lambda \sum_{i=1}^{n}y_i$$
so i sum the y's to get 32.8, let T equal zero and n = 4
$$\lambda(32.8) = \frac{4}{\lambda}$$
$$\lambda^2 = \frac{4}{32.8}$$
$$\lambda = \sqrt{\frac{4}{32.8}}$$
$$\lambda = 0.349$$
I have definitely got those four sample values right. I have done this sum three times now. Can anybody see where i'm going wrong?
Your derivative is wrong. The sum should not have a $\lambda$ next to it. The correct loglikelihood derivative is
$$\frac{dL}{d\lambda} = \frac{n}{\lambda} - \sum y_i$$
setting equal to zero then yeilds
$$\hat{\lambda} = \frac{n}{\sum y_i} =\frac{1}{\bar{y}}$$
Plugging in your values yields $$\hat{\lambda} = \frac{1}{8.2} \approx 0.122$$