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Consider two independent random variables $X$ and $Y$.

Let both PDFs are defined by $$f_X(x)=\begin{cases}\displaystyle\frac{4-2x}{3}&\text{if }x\in(0,1),\\0&\text{otherwise.}\end{cases}$$

Find $\Pr(Y \le 2X)$.


I solved this the following process.

  1. I've found $$ \Pr(Y \le 2X | X=c) = \begin{cases} 0 & \text{if }c\le0\\ -\frac{4}{3}(c^2-2c) & \text{if } 0\lt c \lt \frac12\\ 1 & \text{if }c\ge\frac12 \end{cases} $$
  2. I've solved \begin{align} \Pr(Y \le 2X) &= \int_{-\infty}^{\infty} \Pr(Y\le 2X|X=x) f_X(x)~dx\\ &= \int_{0}^{1} \Pr(Y\le 2X|X=x) \cdot \frac{4-2x}{3} ~dx\\ &= \int_{0}^{\frac12} -\frac{4}{3}(x^2-2x) \cdot \frac{4-2x}{3} ~dx + \int_{\frac12}^1 1 \cdot \frac{4-2x}{3} ~dx\\ &= \frac{121}{216} \end{align}

Is there anything wrong? Also, is there the better solution? Thank you for reading my question.

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Yes, indeed, that's the right integral. But not the right value. $$\begin{align}\mathsf P(Y\leq 2X) ~=~& \int_0^{1/2}\frac{4-2x}3\int_{0}^{2x}\frac{4-2y}3\,\mathsf d y\,\mathsf d x +\int_{1/2}^1\frac{4-2x}3\int_0^1 \frac{4-2y}3\,\mathsf d y\,\mathsf d x \\[1ex] ~=~& \int_0^{1/2}\frac{4-2x}3\frac{4(2x-x^2)}3\,\mathsf d x +\int_{1/2}^1\frac{4-2x}3\mathsf d x \\[1ex] ~=~& \int_0^{1/2}\frac{32x-32x^2+8x^3}9\,\mathsf d x +\int_{1/2}^1\frac{4-2x}3\mathsf d x \\[1ex]=~& \frac 89\left[\frac{x^4}4-\frac{4x^3}3+2x^2\right]_{x=0}^{x=1/2}+\frac 13\left[4x-x^2\right]_{x=1/2}^{x=1} \\[2ex]=~& \dfrac{157}{261} \end{align}$$

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