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Here are my questions:

  1. Let $X$ ~ Unif$(0, 1)$, and $0<a<b<1$. Also, let \begin{cases} Y = 1 & \text{if $0<X<b$} \\ Y = 0 & \text{otherwise} \\ \end{cases}

Similarly, let \begin{cases} Z = 1 & \text{if $a<X<1$} \\ Z = 0 & \text{otherwise} \\ \end{cases}

Find the variance-covariance matrix of Y, Z.

What I tried: Note that P($0<X<b$) = $F_X(b)$ - $F_X(0)$ = $\frac{x}{b}$.

Similarly, note P($a<X<1$) = $F_X(1)$ - $F_X(0)$ = $\frac{x-a}{x-b}$.

Then, $E(Y)$ = $1*P($0<X<b$)$ + $0*[$1$-P($0<X<b$)]$ = $\frac{x}{b}$. Similarly, E(Z) = $\frac{x-a}{1-a}$. Here, I get stuck (and I don't even know if it's right).

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  • $\begingroup$ There are two not-so-short questions here. Please ask (2) in a separate thread. $\endgroup$ – gunes Jun 21 at 18:28
  • $\begingroup$ Where specifically are you having a problem? I'm not really sure on this one. $\endgroup$ – wolfies Jun 21 at 19:25
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    $\begingroup$ @BoJack does the answer help you in any way? $\endgroup$ – gunes Jun 23 at 15:47
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    $\begingroup$ They seem correct. $\endgroup$ – gunes Jun 23 at 19:57
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    $\begingroup$ Sounds good. I've done so! $\endgroup$ – Bo Jack Jun 23 at 20:14
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First of all, $P(0<X<b)=b$ not $x/b$, i.e. not a function of $x$. Similarly, $P(a<X<1)=1-a$. These are also equal to $E[Y],E[Z]$ respectively. Also, we have $E[Y^2]=E[Y],E[Z^2]=E[Z]$ for binary RV case. For the joint moment, $E[YZ]=P(Y=1\cap Z=1)=P(a<X<b)=b-a$. I think you can follow from here.

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