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A random variable $Y \sim U[0,1]$. Let $W = \frac7{10} - b\cdot(\frac7{10} - y)^2, 0<b<1$.
Completely specify the CDF and PDF of $W$.
Also show that the PDF of W integrates to $1$.

So I have worked through most of this problem. Importantly, the transformation is $2:1$ on the interval y $\in (\frac25,1]$ and $1:1$ on the interval y $\in [0, \frac25)$. Thus I think the CDF is:

$$\begin{cases} \frac3{10}+\sqrt{\frac{7/10 - w}b } & \text{for}~~ w \in (\frac7{10} - \frac{49b}{100} ,\, \frac7{10} - \frac{9b}{100}) \\ 2\sqrt{ \frac{7/10 - w}b } & \text{for}~~ w \in (\frac7{10} - \frac{9b}{100} ,\, \frac7{10}) \\ \end{cases}$$

Can anyone confirm this? If I know this is right, I can figure out the rest on my own. When integrating the PDF, I am running into some strange (${}+{}$ or ${}-{}$) situations that integrate to $1$ only if I select a certain sign, so I am having trouble verifying the CDF is correct by integrating the PDF over its support.

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That is not quite what I get...

$$\begin{align}F_W(w) &=\mathsf P(0.7 -b(0.7-Y)^2\leqslant w) \\[1ex]&=\mathsf P(Y\leq 0.7-\sqrt{\tfrac{0.7-w}b})+\mathsf P (Y\geq 0.7+\sqrt{\tfrac{0.7-w}b}) \\[1ex]&=(0.7-\sqrt{\tfrac{0.7-w}b})\mathbf 1_{0\leq (0.7-\sqrt{\tfrac{0.7-w}b})\leq 1}+(0.3-\sqrt{\tfrac{0.7-w}b})\mathbf 1_{0\leq (0.7+\sqrt{\tfrac{0.7-w}b})\leq 1}+\mathbf 1_{0.7<w} \\[1ex]&=(0.7-\sqrt{\tfrac{0.7-w}b})\mathbf 1_{0.7-0.49b\leq w\leq 0.7}+(0.3-\sqrt{\tfrac{0.7-w}b})\mathbf 1_{0.7-0.09b\leq w\leq 0.7}+\mathbf 1_{0.7\lt w} \\[1ex]&=(0.7-\sqrt{\tfrac{0.7-w}b})\mathbf 1_{0.7-0.49b\leq w\lt 0.7-0.09b}+(1-2\sqrt{\tfrac{0.7-w}b})\mathbf 1_{0.7-0.09b\leq w\lt 0.7}+\mathbf 1_{0.7\leq w} \\[3ex] F_W(w) &= \begin{cases}0&:& \quad w\lt 0.7-0.49b\\0.7-\sqrt{(0.7-w)/b~}&:&0.7-\sqrt{(0.7-w)/b~}\leqslant w\lt 0.7-0.09b\\1-2\sqrt{(0.7-w)/b~}&:& 0.7-0.09b\leqslant w< 0.7\\1&:& 0.7\leqslant w \end{cases} \end{align}$$

To check, I have $F_W([0.7-0.49b]^-)=0$ and $F_W([0.7]^+)=1$ as we require.

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