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The distribution of values of the retirement package offered by a company to new employees is modeled by the probability density function:
$$ f(x) = \begin{cases} \frac 1 5 e^{-\frac 1 5 (x-5)}, &x>5 \\ 0, &{\rm otherwise.} \end{cases} $$

Calculate the variance of the retirement package value for a new employee, given that the value is at least $10$.

Part of Solution:
$$ f(x) = \frac{0.2e^{-0.2(y-5)}}{\int_{10}^\infty 0.2e^{-0.2(y-5)}dx} = \frac{0.2e^{-0.2(y-5)}}{-e^{-0.2(y-5)}|_{10}^\infty} = \frac{0.2e^{-0.2(y-5)}}{e^{-0.2(5)}} = 0.2e^{-0.2(y-10)},\\ y>10. $$

Then note that $Y – 10$ has an exponential distraction with mean $5$. Subtracting a constant does not change the variance, so the variance of $Y$ is also $25$.

I don't know what formula they are using to get $f(y)$ in this case. Can someone tell me what formula it is?

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  • $\begingroup$ This is a conditional distribution. Actuarial Exams? $\endgroup$ – Matthew Drury Jan 10 '18 at 1:07
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Hint: Suppose $X$, the retirement package payment, has density function $f(x)$ given above.

Then, given that the value is at least 10, the new PDF would be $$f_{X \mid X > 10}(x)=\dfrac{f(x)}{\mathbb{P}(X > 10)}\text{.}$$ You can think of this as a consequence of the conditional probability formula: $$\mathbb{P}(A \mid B)=\dfrac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}\text{.}$$

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This is a conditional density, the density of $X$ conditional on the event that $X>10$: start with $$\mathbb{P}(X\in B\,,\ X>10)=\int_{B\cap (10,\infty)} 0.2 e^{-0.2(x-5)}\text{d}x$$ and$$\mathbb{P}(X>10)=\int_{10}^\infty 0.2 e^{-0.2(x-5)}\text{d}x$$

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