My Math Forum correlation coefficient

 September 20th, 2011, 04:15 PM #1 Newbie   Joined: Sep 2011 Posts: 2 Thanks: 0 correlation coefficient So here's the thing... I think I only need to know how to get the expected value of two dependent variables $E[U*V]$, but just in case I'm wrong I'll give you all the data. Let $X, Y, Z$ be independent random variables with uniform distribution over a closed interval $[-a,a]$. Then we define variables $U$ and $V$. $X,Y,Z\sim U[-a,a] \\ U:=min\{X,Y,Z\} \\ V:=max\{X,Y,Z\} \\[3mm]$ We're looking for a corelation coeficient between $U$ and $V$, where this is its formulae: $\rho_{U,V}=\frac{E(U*V)-E(U)*E(V)}{\sqrt{D(U)*D(V)}}$ $E(X)$ and $D(X)$ are expected value and dispersion respectively for a variable $X$. For $u\in [-a,a]$ is $F_U$ the cumulative distribution function of a variable $U$: $F_U(u)=P(U\leq u)=P(min\{X,Y,Z\}\leq u)=1-P(min\{X,Y,Z\}>u)= \\ =1-P(X>u,Y>u,Z>u)\stackrel{indep.}{=}1-P(X>u)P(Y>u)P(Z>u)= \\ =1-P(X>u)^3=1-(\frac{a-u}{2a})^3 \\[3mm]$ and $f_U$ probability density: $f_U(u)=F#39;_U(u)=\frac{3(a-u)^2}{8a^3} \\[3mm]$ So we can get the expected value of $U$: $E(U)=\int_{-a}^au f_U(u) du=\int_{-a}^a\frac{3}{8a^3}(a^2-2au+u^3)du=\frac{3}{8a^3}(\frac{a^2u^2}{2}-\frac{2au^3}{3}+\frac{u^4}{4})|_{-a}^a= \\ =-\frac{3*4a^4}{8a^3*3}=-\frac{a}{2}$ Analogous for $V$: $v\in [-a,a] \\[2mm] F_V(v)=P(V\leq v)=P(max\{X,Y,Z\}\leq v)=P(X\leq v,Y\leq v,Z \leq v)\stackrel{indep.}{=} \\ \stackrel{indep.}{=}P(X\leq v)P(Y\leq v)P(z\leq v)=P(X\leq v)^3=(\frac{a+v}{2a})^3 \\[3mm] f_V(v)=F'_V(v)=\frac{3(a+v)^2}{8a^3} \\[3mm] E(V)=\int_{-a}^a v f_V(v)dv=\cdots =\frac{a}{2} \\[2mm]$ As expected due to the simetry of $U$ and $V$ over $[-a,a]$. \\ Now we need the expected value of the product of $U&V$: $W:=U*V \\[3mm] F_W(w)=P(W\leq w)=P(U V\leq w)=P(min\{X,Y,Z\}*max\{X,Y,Z\}\leq w)=? \\[3mm] f_W(w)=F'_W(w)=?' \\[3mm] E(W)=\int_{-a^2}^{a^2}w f_W(w)dw=\cdots \\$ Now the dispersions: $F_{U^2}(u)=P(U^2\leq u)=P(min\{X,Y,Z\}^2\leq u)=1-P(min\{X,Y,Z\}^2>u)=??? \\[3mm] E(U^2)=\int_{0}^{a^2}\cdots \\[3mm] D(U)=E(U^2)-E(U)^2=(*) \\$ I thank you in advance for all your help, matmul
 October 24th, 2011, 08:11 AM #2 Newbie   Joined: Sep 2011 Posts: 2 Thanks: 0 Re: correlation coefficient i suppose noone knows the answer?

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