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If the probability density function is ($0\le x \le 1, 0\le y \le1$):

(i) $f_{X}(x) = \frac{3x^{2}}{2} + x$

(ii) $f_{Y}(y) = \frac{3y^{2}}{2} + y$

Find the distribution functions $F_{X}(x) = P(X\le x)$ and $F_{Y}(y) = P(Y\le y)$.

Can someone check these are correct:

(i) $F_{X}(x) = \frac{x^{3}}{2} + \frac{x^{2}}{2}$

(ii)$F_{Y}(y) = \frac{y^{2}}{2} + \frac{y^{3}}{2}$

user2850514
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1 Answers1

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$(0≤x,y≤1)$

$$F_X(x) = \int_{0}^x f_X(z)\;\mathrm{d}z=\int_{0}^x \dfrac{3z^2}{2}+z \;\mathrm{d}z= \dfrac{x^3}{2}+\dfrac{x^2}{2}$$

$$F_Y(y) = 1-\int_{y}^{1} f_Y(z)\;\mathrm{d}z=1-\int_{y}^1 \dfrac{3z^2}{2}+z \;\mathrm{d}z= 1 - (1- \dfrac{y^3}{2}-\dfrac{y^2}{2}) = \dfrac{y^3}{2}+\dfrac{y^2}{2}$$

Graham Kemp
  • 129,094