excans:98102309ac: Difference between revisions
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'''Solution: B''' | |||
Define <math>g(x) = kx </math> and <math>h(y) = 1 </math>. Then <math>f(x,y) = g(x) h(x) </math>. In other words, <math>f(x,y)</math> can be written as the product of a function of <math>x</math> alone and a function of <math>y</math> alone. It follows that <math>X</math> and <math>Y</math> are independent. Therefore, <math>\operatorname{Cov}[X, Y] = 0.</math> | Define <math>g(x) = kx </math> and <math>h(y) = 1 </math>. Then <math>f(x,y) = g(x) h(x) </math>. In other words, <math>f(x,y)</math> can be written as the product of a function of <math>x</math> alone and a function of <math>y</math> alone. It follows that <math>X</math> and <math>Y</math> are independent. Therefore, <math>\operatorname{Cov}[X, Y] = 0.</math> | ||
{{soacopyright | 2023}} | {{soacopyright | 2023}} | ||
Latest revision as of 14:44, 7 May 2023
Solution: B
Define [math]g(x) = kx [/math] and [math]h(y) = 1 [/math]. Then [math]f(x,y) = g(x) h(x) [/math]. In other words, [math]f(x,y)[/math] can be written as the product of a function of [math]x[/math] alone and a function of [math]y[/math] alone. It follows that [math]X[/math] and [math]Y[/math] are independent. Therefore, [math]\operatorname{Cov}[X, Y] = 0.[/math]
Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.