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Let Y be a random variable denoting the age at which a piece of equipment fails. In reliability theory, the probability that an item fails at time y given that it has survived until time y is called the hazard rate, $h(y)$. In terms of the pdf and cdf,

$$h(y) = \frac{f_Y(y)}{1 - F_Y(y)}$$

Find $h(y)$ if Y has an exponential pdf.

So I set $f_Y(y)$ to the exponential function

$$f_Y(y) = \lambda e^{-\lambda y}$$

And

$$F_Y(y) = \int \lambda e^{-\lambda y} = -e^{-\lambda y}$$

Therefore

$$h(y) = \frac{\lambda e^{-\lambda y}}{1 + e^{-\lambda y}}$$

My question is: Am I right thus far, and if so, can I take it any further? Right now I don't see any way to reduce it further.

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Your CDF is wrong, it should be $F_Y(y) = 1-e^{-\lambda y}$, and then you can reduce further your answer.


For an exponential RV, you have indeed $f_Y(t) = \lambda e^{-\lambda t}$, and then the CDF is given by $F_Y(y) = \int_{0}^y \lambda e^{-\lambda t}$. You probably forgot the contribution of the lower bound of the integral.

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You made a mistake in calculating the CDF. The CDF defined as

$$F_{Y}(y) = \int_{-\infty}^{y}f_{Y}(t)dt$$

Since $f_{Y}(y) = 0$ for $y < 0$ the CDF of the exponential distribution is

$$F_{Y}(y) = \int_{0}^{y}\lambda e^{-\lambda t} dt = 1 - e^{-\lambda y}$$

Then

$$h(y) = \frac{f_{Y}(y)}{1 - F_{Y}(y)} = \frac{\lambda e^{-\lambda y}}{e^{-\lambda y}} = \lambda$$

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  • $\begingroup$ Your integral should go from $0$ to $y$: the PDF of the exponential law is $0$ for negative numbers and the integral you wrote diverges. $\endgroup$ – Dr_Sam Oct 23 '14 at 6:29

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