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I'm working through some homework problems in Hajeck. I'm trying to understand how to find the CDF given a random variable description. The problem is here:

(Hajek 1.14) CDF and characteristic function of a mixed type random variable

Let $X = (U - 0.5)_+$, where $U$ is uniformly distributed over the interval $[0, 1]$. That is, $X = U - 0.5$ if $U - 0.5 \geq 0$, and $X = 0$ if $U - 0.5 < 0$. Find and carefully sketch the CDF $F_X$ . In particular, what is $F_X(0)$?

I can write the following description:

\begin{equation} F(x) = \begin{cases} 0 & U < 0 \\ U-0.5 & U \geq 0.5 \\ \end{cases} \end{equation}

But what do I do with U? It seems like I should calculate some kind of value or formula for U. I know it's uniform. So

\begin{equation} F(x) = \begin{cases} 1 & x \geq b\\ \frac{x-a}{b-a} & a \leq x < b\\ 0 & x < a \end{cases} \end{equation}

Edit: I think this formula is applicable here:

\begin{equation} \begin{split} F_X(x) = P(U-0.5 < 0)P(X\leq x\mid U < 0.5) + P(U-0.5 \geq 0)P(X\leq x \mid U\geq 0.5) \end{split} \end{equation}

Because $U$ is normal on $[0, 1]$

\begin{equation} \begin{split} P(U<0.5) = \frac{1}{2}\\ P(U\geq0.5) = \frac{1}{2} \end{split} \end{equation}

When $U<0.5$, $X=0$. Given that $P<0.5$ it must be that $P(X\leq x\mid U<0.5)=1$ and that $P(X\leq x\mid U\geq 0.5) = U-0.5$. Since $U$ is uniform over $[0,1]$, $F_U(u) = x-1$ in the interval given.

I'm still confused about how to deal with $U - 0.5$ and $F_U(u)$. I'd like to say that the $P(X\leq x\mid U\geq 0.5) = x-1$ (because $U$ is uniform) but I'm a bit uncertain about the truth of that.

Edit 2: I don't think that above edit works out correctly, because $P(X)\neq 1$. So I'm missing something here.

FirmwareRootkits
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1 Answers1

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The formula $$F(x)=\left\{\begin{array}{ll} 0 & : U < 0 \\ U-.5 & : U \geqslant .5\end{array}\right.$$ should say $$X=\left\{\begin{array}{ll} 0 & : U < 0 \\ U-.5 & : U \geqslant .5\end{array}\right.$$ This is a formula for $X$, not $F(x)$.

I would use this formula to directly evaluate $\mathbb{P}(X\leqslant t)$ for different values of $t$. This will look different for $t<0$, $0\leqslant t < .5$, and $t\geqslant .5$. For example, $\mathbb{P}(X\leqslant t)=0$ for $t<0$, since $X$ is always at least $0$. $\mathbb{P}(X\leqslant t)=1$ for $t\geqslant .5$, since $U-.5$ is at most $.5$. For $0\leqslant t<.5$, we can write $$\mathbb{P}(X\leqslant t) = \mathbb{P}(X=0)+\mathbb{P}(0<X\leqslant t) = \ldots,$$ where $\ldots$ is filled in using your piecewise formula to obtain events in terms of $U$ rather than $X$. Once the events are in terms of $U$, we can evaluate their probabilities.

user469053
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