A calculus problem about statistics

Probability Level 4

The random variable ( X , Y ) (X, Y) has a density function : f ( x , y ) = { k y e − x if 0 < x < y < 2 x 0 for all other points f(x,y) = \begin{cases} kye^{-x} & \text {if } 0 < x < y < 2x \\ 0 & \text {for all other points} \end{cases}

Find k k

Submit ⌊ 2000 k ⌋ \lfloor 2000k \rfloor .

Notation: ⌊ ⋅ ⌋ \lfloor \cdot \rfloor denotes the floor function .


The answer is 666.

Guillermo Templado by Guillermo Templado , 46, Spain

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1 solution

1 = ∫ ∫ R 2 f ( x , y ) d x d y = ∫ 0 ∞ ∫ x 2 x k y e − x d y d x = 1 = \displaystyle \int \int_{\mathbb{R}^2} f(x,y) dx \space dy = \int_0^{\infty} \int_x^{2x} kye^{-x} dy \space dx = = k ⋅ ∫ 0 ∞ e − x [ y 2 2 ] x 2 x d x = 3 k 2 ⋅ ∫ 0 ∞ x 2 ⋅ e − x d x = 3 k 2 ⋅ Γ [ 3 ] = 3 k ⇒ = k\cdot \int_0^{\infty} e^{-x}\left[\frac{y^2}{2}\right]_x^{2x} dx = \frac{3k}{2}\cdot \int_0^{\infty} x^2\cdot e^{-x} dx = \frac{3k}{2}\cdot \Gamma[3] = 3k \Rightarrow ( Γ [ 3 ] = 2 ! = 2 \color{#D61F06}{\Gamma[3] = 2! = 2} ) 1 = 3 k ⇒ k = 1 3 ⇒ ⌊ 2000 k ⌋ = 666 1 = 3k \Rightarrow k = \frac{1}{3} \Rightarrow \lfloor 2000k \rfloor = 666

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