Random variable X and Y have the joint PDF fXY(x,y)={c, x≥0,y≥0,(x2+y)≤1,0, otherwise Find the constant c.

The joint density function of two random variables X and Y is given below: fXY(x,y)={c, x≥0,y≥0,(x2+y)≤1,0, otherwise c∫01∫01−x2dxdy=1 ∫01∫01−x2dxdy=1c ∫01[∫01−x2dy]dx=1c ∫01[y]01−x2dx=1c ∫01(1−x2)dx=1c [x]01−[x33]01=1c (1)−(13)=1c 23=1c c=32 Thus, the value of the constant c is 32.

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Wierzycaz

Answered question

2021-10-23

Random variable X and Y have the joint PDF
$f_{X Y} (x, y) = {\begin{cases} c, x \geq 0, y \geq 0, (x^{2} + y) \leq 1, \\ 0, o t h e r w i s e \end{cases}$
Find the constant c.

Answer & Explanation

wornoutwomanC

Skilled2021-10-24Added 81 answers

The joint density function of two random variables X and Y is given below:
$f_{X Y} (x, y) = {\begin{cases} c, x \geq 0, y \geq 0, (x^{2} + y) \leq 1, \\ 0, o t h e r w i s e \end{cases}$
$c \int_{0}^{1} \int_{0}^{1 - x^{2}} d x d y = 1$
$\int_{0}^{1} \int_{0}^{1 - x^{2}} d x d y = \frac{1}{c}$
$\int_{0}^{1} [\int_{0}^{1 - x^{2}} d y] d x = \frac{1}{c}$
$\int_{0}^{1} {[y]}_{0}^{1 - x^{2}} d x = \frac{1}{c}$
$\int_{0}^{1} (1 - x^{2}) d x = \frac{1}{c}$
${[x]}_{0}^{1} - {[\frac{x^{3}}{3}]}_{0}^{1} = \frac{1}{c}$
$(1) - (\frac{1}{3}) = \frac{1}{c}$
$\frac{2}{3} = \frac{1}{c}$
$c = \frac{3}{2}$
Thus, the value of the constant c is $\frac{3}{2}$ .

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