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In the following example for EM

http://math.usask.ca/~longhai/teaching/stat812-1409/rdemo/EM_examples.pdf

The equation for the expected value of a light bulb that has burned out is introduced:

$$ \theta - \frac{te^{-t/\theta ^{(k)}}}{1-e^{-t/\theta ^{(k)}} } $$

How was this equation derived?

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    $\begingroup$ Please state the example fully here in your question so that we do not have to refer to material on another site to understand it. $\endgroup$ – whuber May 15 '17 at 21:04
  • $\begingroup$ It seems to me that it derives from the E[U] = sum(U*Pr{U < t}) $\endgroup$ – Guilherme Marthe May 15 '17 at 21:06
  • $\begingroup$ Typically the value of a burned out lightbulb would be taken to be approximately 0. People throw them away. Do you mean the expected age of a just-burned-out lightbulb? Also as whuber asks, please include more details from your link so that your question can be understood even if the link is moved or the paper is taken down. $\endgroup$ – Glen_b -Reinstate Monica May 16 '17 at 4:26
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The expression is for $E[X|X<t],$ where $X$ has an exponential distribution with mean $\theta$ and $t>0.$ You can find it using this formula: $$E \left[ X | X < t \right] = \frac{\int_0^t xf(x) dx}{P \left[ X<t \right]}$$

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