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So I have this PDF: $$ f(x)= \begin{cases} x + 3 & \text{ for } -3 \leq x < -2\\ 3 - x & \text{ for } 2 \leq x < 3\\ 0 & \text{ otherwise} \end{cases} $$

To make this a CDF, I have integrated the PDF from $-\infty$ to some value, $x$.

$$ F(x)= \begin{cases} \frac{x^2}{2} + 3x + \frac{9}{2} & \text{ for } -3 \leq x <-2\\ \frac{1}{2} & \text{ for } -2 \leq x<2\\ \frac{-x^2}{2} + 3x + \frac{7}{2} & \text{ for } 2 \leq x<3 \end{cases} $$

My friend argues that the first term in this CDF which is $(x^2/2 + 3x + 9/2)$ should actually be $(x^2/2 + 3x)$. But isn't this impossible? At $x = -3$, the CDF must be $0$, am I correct?. This is only true in the case where the first term is $(x^2/2 + 3x + 9/2)$.

If someone could shed light on this topic, that would be much appreciated.

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    $\begingroup$ You don't integrate all the way to $\infty$ you integrate up to a fixed (but arbitrary) value, $y$. $\endgroup$ Commented Jun 30, 2014 at 3:57
  • $\begingroup$ yes, I have done just that for my first two terms. can someone confirm, or discuss this question further? $\endgroup$
    – baba
    Commented Jun 30, 2014 at 4:00
  • $\begingroup$ I've upgraded my comment to an answer below. $\endgroup$ Commented Jun 30, 2014 at 4:10

6 Answers 6

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$ f(x) = \begin{cases} 0 &, \phantom{-3 \leq {}}x< -3 \\ x+3 &, -3 \leq x \leq -2 \\ 0 &, -2 \leq x \leq \phantom{-{}}2 \\ 3-x &, \phantom{-{}}2 \leq x \leq \phantom{-{}}3 \\ 0 &, \phantom{-{}}3 \leq x\end{cases}$

\begin{align} F(x) = \int_{-\infty}^x f(t) \,\mathrm{d}t &= \begin{cases} 0 &, \phantom{-3 \leq {}}x< -3 \\ 0+\int_{-3}^x t+3 \,\mathrm{d}t &, -3 \leq x \leq -2 \\ 0+\int_{-3}^{-2} t+3 \,\mathrm{d}t + 0&, -2 \leq x \leq \phantom{-{}}2 \\ 0+\int_{-3}^{-2} t+3 \,\mathrm{d}t + \int_{2}^x 3-t \,\mathrm{d}t&, \phantom{-{}}2 \leq x \leq \phantom{-{}}3 \\ 0+\int_{-3}^{-2} t+3 \,\mathrm{d}t + \int_{2}^3 3-t \,\mathrm{d}t + 0 &, \phantom{-{}}3 \leq x\end{cases} \\ &= \begin{cases} 0 &, \phantom{-3 \leq {}}x< -3 \\ 0+\frac{x^2+6x+9}{2} &, -3 \leq x \leq -2 \\ 0+\frac{1}{2} &, -2 \leq x \leq \phantom{-{}}2 \\ 0+\frac{1}{2} + \frac{-x^2+6x-8}{2}&, \phantom{-{}}2 \leq x \leq \phantom{-{}}3 \\ 0+\frac{1}{2} + \frac{1}{2} + 0 &, \phantom{-{}}3 \leq x\end{cases} \end{align}

As you can see $F(-3) = 0$ because $\frac{9-18+9}{2} = 0$. Your argument is correct that the CDF must increase from zero (starting) at $-3$.

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  • $\begingroup$ I see that the OP's $f$ has changed... Editing. $\endgroup$ Commented Jun 30, 2014 at 4:23
  • $\begingroup$ 3-x for 2≤x≤3, not −2≤x≤3 as you have stated. $\endgroup$
    – baba
    Commented Jun 30, 2014 at 4:26
  • $\begingroup$ @baba: So in the three minutes from my comment and start to re-edit until your comment, you didn't bother to read my comment? Also, updated for the changed OP. $\endgroup$ Commented Jun 30, 2014 at 4:31
  • $\begingroup$ Oh, I'm really sorry. I wasn't paying attention. Thank you for the answer, it is well written. $\endgroup$
    – baba
    Commented Jun 30, 2014 at 4:36
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Strictly speaking the CDF should be defined for all $x\in\Bbb R$. If $x\le-3$ we have $$F(x)=\int_{-\infty}^x f(t)\,dt=\int_{-\infty}^x 0\,dt=0\ .$$ If $-3<x<-2$ then $$F(x)=\int_{-\infty}^x f(t)\,dt =\int_{-\infty}^{-3} 0\,dt+\int_{-3}^x t+3\,dt =\frac{1}{2}(x+3)^2$$ (which in fact is the answer you have already, but it's simpler this way). For $-2\le x\le2$ we get $F(x)=\frac{1}{2}$ as you have already. For $2<x<3$ we obtain $$F(x)=\int_{-\infty}^x f(t)\,dt =\int_{-\infty}^2 f(t)\,dt+\int_2^x f(t)\,dt =\frac{1}{2}+\int_2^x (3-t)\,dt=1-\frac{1}{2}(3-x)^2\ .$$ Finally, if $x\ge 3$ then $F(x)=1$.

As a check, note that you should have $F(x)$ always increasing and $$\lim_{x\to-\infty} F(x)=0\quad\hbox{and}\quad \lim_{x\to\infty} F(x)=1\ .$$

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The cdf $F(x)$ is given by $$F(x) =\begin{cases} 0 & \text{ for } x <-3\\ \int_{-3}^{x}(t+3) \, dt & \text{ for } -3 \leq x <-2\\ \frac{1}{2} & \text{ for } -2 \leq x <2\\ \frac{1}{2}+\int_{-2}^{x}(3-t) \, dt & \text{ for } 2 \leq x <3\\ 1 & \text{ otherwsie} \end{cases} $$ My suggestion is to plot the pdf $f(x)$ and then think in terms of the area of triangle(s) being swept as you move along the $x-$axis.

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Hint: integrate $ 3-x$. Say, integral is $f(x)$. Plug in $f(3)=1$ to get the value of constant

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Let's put it this way

$$\int_{-\infty}^Y f(x)= \begin{cases} 0 & -\infty < Y < -3 \\ Y^2+3Y+{9\over 2} & -3\le Y <-2 \\ {1\over 2} & -2\le Y <2 \\ {1\over 2} +\int_2^Y(3-x)\, dx & 2\le Y < 3 \\ 1 & Y\ge 3 \end{cases}$$

The first term is INDEED including the 9/2, this is because it is

$$\int_{-3}^Y f(x)\,dx\int_{-3}^Y (x+3)\,dx$$

By the fundamental theorem of calculus, this is the antiderivative at the top limit, Y, minus the antiderivative at the bottom limit, -3. The value of $x^2/2+3x$ at $x=-3$ is $-9/2$ so you get the result with the +9/2 because you subtract the bottom limit and subtracting a negative gives a positive.

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Note first of all that the cdf is defined for all $x$, so a complete answer should take care of all $x$.

If $x\lt -3$, everything is easy. We have $\int_{-\infty}^x f_X(t)\,dt=0$. So $F_X(x)=0$ if $x\lt -3$.

Now for $-3\le x\lt -2$, in effect we are integrating $t+3$ from $-3$ to $x$, since there is no "mass" before $-3$. An antiderivative is $t^2/2+3t$. Plug in. We get $\frac{x^2}{2}+3x+\frac{9}{2}$.

Note that this is $\frac{1}{2}$ at $x=-2$. The density function is then $0$ for a while, so exactly as you had it, if $-2\le x\lt 2$ then $F_X(x)=\frac{1}{2}$.

Now for $2\le x\lt 3$, we want the integral up to $2$, plus the integral from $2$ to $x$ of $3-t$. We get $$\frac{1}{2}+(3x-\frac{x^2}{2})- 4.$$ Finally, for $x\ge 3$, the cdf is $1$.

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