exercise:B8ee1f5c45: Difference between revisions
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<li><math>(0.1y )^{1.25} \exp[−0.125(0.1y)^{0.25} ]</math></li> | <li><math>(0.1y )^{1.25} \exp[−0.125(0.1y)^{0.25} ]</math></li> | ||
<li><math>0.125(0.1y )^{0.25} \exp[−(0.1y )^{1.25} ]</math></li> | <li><math>0.125(0.1y )^{0.25} \exp[−(0.1y )^{1.25} ]</math></li> | ||
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{{soacopyright | 2023}} | {{soacopyright | 2023}} | ||
Latest revision as of 19:59, 4 May 2023
An actuary models the lifetime of a device using the random variable [math]Y = 10X^{0.8}[/math], where [math]X[/math] is an exponential random variable with mean 1. Let [math]f(y)[/math] be the density function for [math]Y[/math].
Determine [math]f(y)[/math] for [math]y\gt0[/math].
- [math]10 y^{0.8} \exp(−8 y^{−0.2} )[/math]
- [math]8 y^{−0.2} \exp(−10 y^{0.8} )[/math]
- [math]8 y^{−0.2} \exp[−(0.1y)^{1.25} ][/math]
- [math](0.1y )^{1.25} \exp[−0.125(0.1y)^{0.25} ][/math]
- [math]0.125(0.1y )^{0.25} \exp[−(0.1y )^{1.25} ][/math]
Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.