Question

The random variables $Y_1$ and $Y_2$, for $-1 \le \alpha \le 1$, have the joint density function given by $\begin{cases} \left[1 - \alpha \left( (1 - 2e^{-Y_1})(1 - 2e^{-Y_2}) \right) \right] e^{-Y_1 - Y_2}, & 0 \le Y_1, 0 \le Y_2 \\ 0, & \text{elsewhere} \end{cases}$ It can be established that the marginal distributions of $Y_1$ and $Y_2$ are both exponential with mean 1 and it can be shown that $Y_1$ and $Y_2$ are independent if and only if $\alpha = 0$. (a) Derive $Cov(Y_1, Y_2)$. $Cov(Y_1, Y_2) = $ (b) Show that $Cov(Y_1, Y_2) = 0$ if and only if $\alpha = 0$. If $Cov(Y_1, Y_2) = 0$, then from part (a), $\alpha = $ If $\alpha = 0$, then from part (a), $Cov(Y_1, Y_2) = $ (c) Argue that $Y_1$ and $Y_2$ are independent if and only if $\rho = 0$. First, it can be shown that $Y_1$ and $Y_2$ are independent if and only if $\alpha = 0$. Therefore if $Y_1$ and $Y_2$ are independent, then $\alpha = $ and hence, from part (b), $Cov(Y_1, Y_2) = $ and hence, from part (b), $\alpha = $ and so $\rho = $ If $\rho = 0$, then and so $Y_1$ and $Y_2$ are ----Select----

          The random variables $Y_1$ and $Y_2$, for $-1 \le \alpha \le 1$, have the joint density function given by
$\begin{cases} \left[1 - \alpha \left( (1 - 2e^{-Y_1})(1 - 2e^{-Y_2}) \right) \right] e^{-Y_1 - Y_2}, & 0 \le Y_1, 0 \le Y_2 \\ 0, & \text{elsewhere} \end{cases}$
It can be established that the marginal distributions of $Y_1$ and $Y_2$ are both exponential with mean 1 and it can be shown that $Y_1$ and $Y_2$ are independent if and only if $\alpha = 0$.
(a) Derive $Cov(Y_1, Y_2)$.
$Cov(Y_1, Y_2) = $
(b) Show that $Cov(Y_1, Y_2) = 0$ if and only if $\alpha = 0$.
If $Cov(Y_1, Y_2) = 0$, then from part (a), $\alpha = $
If $\alpha = 0$, then from part (a), $Cov(Y_1, Y_2) = $
(c) Argue that $Y_1$ and $Y_2$ are independent if and only if $\rho = 0$.
First, it can be shown that $Y_1$ and $Y_2$ are independent if and only if $\alpha = 0$. Therefore if $Y_1$ and $Y_2$ are independent, then
$\alpha = $
and hence, from part (b), $Cov(Y_1, Y_2) = $
and hence, from part (b), $\alpha = $
and so $\rho = $
If $\rho = 0$, then
and so $Y_1$ and $Y_2$ are ----Select----

The random variables Y1 and Y2, for -1 ≤α≤ 1, have the joint density function given by
[1 - α( (1 - 2e^-Y1)(1 - 2e^-Y2) ) ] e^-Y1 - Y2, 0 ≤ Y1, 0 ≤ Y2
0, elsewhere
It can be established that the marginal distributions of Y1 and Y2 are both exponential with mean 1 and it can be shown that Y1 and Y2 are independent if and only if α = 0.
(a) Derive Cov(Y1, Y2).
Cov(Y1, Y2) =
(b) Show that Cov(Y1, Y2) = 0 if and only if α = 0.
If Cov(Y1, Y2) = 0, then from part (a), α =
If α = 0, then from part (a), Cov(Y1, Y2) =
(c) Argue that Y1 and Y2 are independent if and only if ρ = 0.
First, it can be shown that Y1 and Y2 are independent if and only if α = 0. Therefore if Y1 and Y2 are independent, then
α =
and hence, from part (b), Cov(Y1, Y2) =
and hence, from part (b), α =
and so ρ =
If ρ = 0, then
and so Y1 and Y2 are —-Select—-

Added by Geoffrey B.

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