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Question: Jack wants to build a wooden cylinder, He decided to choose it's radius (Y) randomly s.t $Y\sim U[0,1]$.

a. What is the probability that the radius is in a closed interval $[\alpha,\beta]$? what is Y's density function?

b. What is the prob. that the volume of the cylinder (with a known height $h$) (X=\Pi Y^2 h) is in a closed interval $[\alpha,\beta]$?what is X's density function?

c. What is E[X]? d. What is Var[X]?

Request I'd like to verify this solution, especially b. (I'm not sure that I'm allowed to do what I did there, and that I can assume X is also uniform. THx

My answers:

5.a. The probability of the RV being in an interval equals the length of the interval, therefore $P(Y\in [\alpha,\beta])=\beta-\alpha$ The density function is $\begin{cases} 1 &Y\in [0,1]\\ 0 & elsewhere \end{cases}$

5.b. $X=\Pi h Y^2$ so $P(a \le X \le b)= P(a \le \Pi h Y^2 \le b)= P(\sqrt \frac {a}{\Pi h} \le Y \le \sqrt \frac{b}{\Pi h})=$ $\sqrt \frac{b}{\Pi h}-\sqrt \frac {a}{\Pi h}$ The passage with the square root is possible due to the fact that all values are positive only. The values of X can be in this interval $[0,\Pi h]$ therefore the density function of X is $\begin{cases} \frac 1{\Pi h} & x\in[0,\Pi h] \\ 0 & elsewhere \end{cases}$

5.c. Lemma (for every uniform RV $\in [0,1]: E[Y^n]=\frac 1{n+1}$- this is proved easily because this is actually $\int _0^1 y^ndy$ We'll be using this now:

$E[X]=E[\Pi h Y^2]=\Pi h E[Y^2]=\frac {\Pi h}3$

5d. $V(X)=V(\Pi h Y^2)=(\Pi h)^2V(Y^2)=(\Pi h)^2[E[Y^4]-E^2[Y^2]]= (\Pi h)^2[\frac 15-\frac19]=\frac{4(\Pi h)^2}{45}$

jreing
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Hint: [for part b) You have the cumulative for $Y$, namely $F(u)=P(Y\le u)=\sqrt{u/(\pi h)},$ if your calculation is right (I didn't check that). The density for $Y$ should be the derivative of this cumulative, which won't be a constant.

coffeemath
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  • The density function: Let $c\in [0,1], F_X=P(X\le c)=P(Y^2h\Pi\le c)=P(Y\le \sqrt{c-\Pi h})=\sqrt{c-\Pi h}$ We know that $f_X=(F_X)'=\frac 1{2\sqrt {c-\Pi h}}$ . Is this correct/not? do I need to add a comment about the function being 0 when c is not in [0,1]? – jreing Nov 20 '13 at 11:16
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    The step from $Y^2h\pi\le c$ to $Y\le \sqrt{c-\pi h}$ is incorrect algebra. You should be dividing through by $h \pi$ there, so it should be $Y \le \sqrt{c/(h \pi)}$ at that step. Take the derivative of that, and yes one does have to say the density is zero outside the relevant interval of $c$, which btw may not be $[0,1]$ since $Y$ has been transformed from $X$. – coffeemath Nov 20 '13 at 11:28