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Let $V \sim f(v) = Tv^2,$ such that $-2<v<1$ and $T$ is a constant. I want to know the distribution of $W=V^2$ and the value $T$. My attempt is that $V = \pm \sqrt{ W }.$ We know that $$\int_{-2}^1 Kx^2 = 1 \quad \Rightarrow 3K =1, \quad K = 1/3$$

Then $V\sim f(v) = \frac{1}{3}v^2, \ -2<v<1$ and $$F_V(v) = \int_{-2}^y \frac{1}{3} v^2 dy + \int_{y}^1 \frac{1}{3} v^2 dy = \frac{y^3+8}{9}+\frac{1-y^3}{9} = 1$$ Then, \begin{align*} F_V (W \le w) & = P(V \le \pm \sqrt{w})\\ & = P(-\sqrt{w} \le V \le \sqrt{w})\\ & = F_V(\sqrt{w}) - F_V(-\sqrt{w})\\ f_V(w) & = \frac12 f_V (w^{-1/2}) - \frac12 f_V (-w^{-1/2})\\ & = \frac12\left(\frac13(w^{-1}) \right) + \frac12\left(\frac13(w^{-1}) \right)\\ & \stackrel{?}{=} \frac{1}{3}w \end{align*}

Could that be wrong?

Ab2020
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1 Answers1

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The value of the constant is right. I continue later with your approach. First of all we need the cdf of $V$, since we work with inequalities. For this purpose we integrate the pdf from $-2$ (lower bound) to $v$:

$$\int_{-2}^v \frac13t^2 \ dt=\left[\frac19t^3\right]_{-2}^v=\frac{v^3}{9}+\frac{8}{9}$$

Thus the cdf is

$$F_V(v)=\begin{cases} 0, \ v\leq -2 \\ \frac{v^3}{9}+\frac{8}{9}, \ -2< v \leq 1 \\ 1, \ v>1 \end{cases}$$

Since the interval is not symmetric around 0 we make a case distinction:

First case: $v\in [-1,1) \Rightarrow w\in [0,1)$. Then we have $v=w^{1/2} $ for $v\in [0,1)$ and $v=-w^{1/2}$ for $v\in [-1,0)$

Second case: $v\in [-2,-1] \Rightarrow w\in [1,4]$

At the first case we can use your interval:

$$F_V(\sqrt{w}) - F_V(-\sqrt{w})=\left(\frac{w^{\frac32}}{9}+\frac{8}{9}\right)-\left(\frac{\left(-w^{1/2}\right)^3}{9}+\frac{8}{9}\right)$$

$$F_W(w)=\left(\frac{w^{\frac32}}{9}+\frac{8}{9}\right)-\left(\frac{-w^{3/2}}{9}+\frac{8}{9}\right)=\frac{w^{\frac32}}{9}+\frac{8}{9}+\frac{w^{\frac32}}{9}-\frac89=\frac{2}{9}w^{\frac32}$$

At the second case $V$ is negative. Thus we have

$P((-V)^2<w)=P(-V<w^{\frac12})=P(V>-w^{\frac12})=1-F\left(-w^{\frac12} \right)$

Inserting the value into the cdf results in

$$F_W(w)=1-\left(\frac{\left(-w^{\frac12}\right)^3}{9}+\frac{8}{9}\right)=1-\left(\frac{-w^{\frac32}}{9}+\frac{8}{9}\right)=\frac{w^{\frac32}}{9}+\frac19$$

callculus42
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  • I'm lost beginning from "...interval is not symmetric", what is rational for considering first and second cases? Wouldn't $F_W(w)$ has bounds? – Ab2020 Feb 04 '23 at 04:16
  • @Ab2020 Your approach works only for sym. Intervals, since the values $-\sqrt w$ and $\sqrt w$ are symetric around $0$. Put in some numbers if it helps. Then for the interval $[-2,-1]$ we cannot bild such a symmetric interval and we have to regard a specific case. Sure the bounds are mentioned at the cases. Btw, with these bounds you can make sanity checks, i.e. $1. F_W(0)=0, \ 2. F_W(1^-)=F_W(1), \ 3. F_W(4)=1$ – callculus42 Feb 04 '23 at 04:36
  • Probably, my problem is my inability to identify when to use the symmetric method and when not to use it. Can you give me some tricks/reference texts for me to be able to differentiate the two methods? – Ab2020 Feb 04 '23 at 04:50
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    You have done it right for the symmetric part. Then we had to regard the part which was left.For all the steps probably some practice is necessary. Here is another exercise which you can make for training. – callculus42 Feb 04 '23 at 05:18
  • I've made some comments (a question) under the first answer: Here but it seems that the user hasn't been back to the website for more than a year. Could you please read my comments and help me with my question over there? Thank you in advance – Ab2020 Feb 05 '23 at 07:30
  • I'm still not able to grasp both 1st and 2nd cases. (1) Generally, how do you pick the intervals? For example, in your 1st case, you pick $v\in [-1,1)$ without saying how/why you pick it, then you came up with $w=[0,1),$ without showing how you had $w.$ (2) You then concluded that $v=w^{1/2}$ for $v\in[0,1)$ and again $v=w^{1/2}$ for $v\in [-1,0).$ (3) I'm struggling with that conclusion too. (4) Almost every textbook that I read do not provide details about how these intervals are obtained. They just assumed the reader knows, which is always a wrong assumption. – Ab2020 Feb 05 '23 at 18:21
  • @Ab2020 I' ve divided the allover interval in a symmetric Intervall and the remaining interval. Quote: "Since the interval is not symmetric around 0 we make a case distinction" I thought rhis was clear. – callculus42 Feb 05 '23 at 18:31
  • I know that $v=[-2,-1,0,1]$ if the split is at 0, do you mean $v_1=[-1,1]$ or $v_1=[-1,1)$? and how was $w=[0,1)?$ – Ab2020 Feb 05 '23 at 18:55
  • @Ab2020 I was not finished with my comment. I hope it clarifies some questions. Ad 2) The split is at $v=0$. For $v\in [0,1)$ we can insert the values $0=0^{1/2} $ and $1=1^{1/2} $ This is true. For the negative part, $v\in [-1,0)$,we must add a negative sign. We cannot calculate a root of a negative value: $v=-w^{1/2}$. Inserting the values of -1 and 0 gives $-1=-1^{1/2}$ and $0=-0^{1/2}$ This is true as well // The distinction between ) and ] is not so imposant. The crucial point is that the interval has no gaps. – callculus42 Feb 05 '23 at 19:06
  • If I understand what you mean about the spilt been at 0 is that you pick the number to the right of 0 to form the interval: $[0, 1]$ and also pick the number to the left of 0 to form: $[-1,0]$? – Ab2020 Feb 05 '23 at 19:20
  • Let us continue this discussion in chat. – Ab2020 Feb 05 '23 at 19:26
  • Yes. If you want to be very precise with the brackets then the second interval is $[-1,0)$. So the intervals do not overlap. – callculus42 Feb 05 '23 at 19:29