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Yves
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As noted by @Yashaswi Mohanty@YashaswiMohanty the expectation of $Y$ can sometimes be found without explicting the probability distribution function.

Assume that all r.vs are of continuous type and $X_1$ is exponential with rate $\lambda >0$. We can consider the survival function $\bar{F}_Y(y) := 1 - F_Y(y)$

$$ \bar{F}_Y(y) = \text{Pr}\{ \min(Z_1,\,Z_2) > y\} =\text{Pr}\{[Z_1 > y] \cap [Z_2 > y] \}. $$ Then by conditioning on $X_1$ we can use the independence \begin{align*} \bar{F}_Y(y) &=\int_0^\infty \text{Pr}\{[Z_1 > y] \cap [Z_2 > y] \, \vert \, X_1 = x_1\} f_{X_1}(x_1) \,\text{d}x_1\\ &= \int_0^\infty \text{Pr}\{[U_1 > y + x_1] \cap [U_2 > y + x_1] \, \vert \, X_1 = x_1\} \, f_{X_1}(x_1) \,\text{d}x_1\\ &= \int_0^\infty \bar{F}_U(y + x_1)^2 \lambda \, e^{-\lambda x_1}\, \text{d}x_1 \end{align*}

There are some cases where we can get a closed form expression. For instance if $U_i$ are exponential with rate $\gamma$ i.e., $\bar{F}_U(u) = e^{-\gamma u}$ for $u >0$.

Interestingly, this is a simple and efficient way to generate a couple of random variables with tail dependence.

As noted by @Yashaswi Mohanty the expectation of $Y$ can sometimes be found without explicting the probability distribution function.

Assume that all r.vs are of continuous type and $X_1$ is exponential with rate $\lambda >0$. We can consider the survival function $\bar{F}_Y(y) := 1 - F_Y(y)$

$$ \bar{F}_Y(y) = \text{Pr}\{ \min(Z_1,\,Z_2) > y\} =\text{Pr}\{[Z_1 > y] \cap [Z_2 > y] \}. $$ Then by conditioning on $X_1$ we can use the independence \begin{align*} \bar{F}_Y(y) &=\int_0^\infty \text{Pr}\{[Z_1 > y] \cap [Z_2 > y] \, \vert \, X_1 = x_1\} f_{X_1}(x_1) \,\text{d}x_1\\ &= \int_0^\infty \text{Pr}\{[U_1 > y + x_1] \cap [U_2 > y + x_1] \, \vert \, X_1 = x_1\} \, f_{X_1}(x_1) \,\text{d}x_1\\ &= \int_0^\infty \bar{F}_U(y + x_1)^2 \lambda \, e^{-\lambda x_1}\, \text{d}x_1 \end{align*}

There are some cases where we can get a closed form expression. For instance if $U_i$ are exponential with rate $\gamma$ i.e., $\bar{F}_U(u) = e^{-\gamma u}$ for $u >0$.

Interestingly, this is a simple and efficient way to generate a couple of random variables with tail dependence.

As noted by @YashaswiMohanty the expectation of $Y$ can sometimes be found without explicting the probability distribution function.

Assume that all r.vs are of continuous type and $X_1$ is exponential with rate $\lambda >0$. We can consider the survival function $\bar{F}_Y(y) := 1 - F_Y(y)$

$$ \bar{F}_Y(y) = \text{Pr}\{ \min(Z_1,\,Z_2) > y\} =\text{Pr}\{[Z_1 > y] \cap [Z_2 > y] \}. $$ Then by conditioning on $X_1$ we can use the independence \begin{align*} \bar{F}_Y(y) &=\int_0^\infty \text{Pr}\{[Z_1 > y] \cap [Z_2 > y] \, \vert \, X_1 = x_1\} f_{X_1}(x_1) \,\text{d}x_1\\ &= \int_0^\infty \text{Pr}\{[U_1 > y + x_1] \cap [U_2 > y + x_1] \, \vert \, X_1 = x_1\} \, f_{X_1}(x_1) \,\text{d}x_1\\ &= \int_0^\infty \bar{F}_U(y + x_1)^2 \lambda \, e^{-\lambda x_1}\, \text{d}x_1 \end{align*}

There are some cases where we can get a closed form expression. For instance if $U_i$ are exponential with rate $\gamma$ i.e., $\bar{F}_U(u) = e^{-\gamma u}$ for $u >0$.

Interestingly, this is a simple and efficient way to generate a couple of random variables with tail dependence.

Source Link
Yves
  • 5.7k
  • 1
  • 23
  • 38

As noted by @Yashaswi Mohanty the expectation of $Y$ can sometimes be found without explicting the probability distribution function.

Assume that all r.vs are of continuous type and $X_1$ is exponential with rate $\lambda >0$. We can consider the survival function $\bar{F}_Y(y) := 1 - F_Y(y)$

$$ \bar{F}_Y(y) = \text{Pr}\{ \min(Z_1,\,Z_2) > y\} =\text{Pr}\{[Z_1 > y] \cap [Z_2 > y] \}. $$ Then by conditioning on $X_1$ we can use the independence \begin{align*} \bar{F}_Y(y) &=\int_0^\infty \text{Pr}\{[Z_1 > y] \cap [Z_2 > y] \, \vert \, X_1 = x_1\} f_{X_1}(x_1) \,\text{d}x_1\\ &= \int_0^\infty \text{Pr}\{[U_1 > y + x_1] \cap [U_2 > y + x_1] \, \vert \, X_1 = x_1\} \, f_{X_1}(x_1) \,\text{d}x_1\\ &= \int_0^\infty \bar{F}_U(y + x_1)^2 \lambda \, e^{-\lambda x_1}\, \text{d}x_1 \end{align*}

There are some cases where we can get a closed form expression. For instance if $U_i$ are exponential with rate $\gamma$ i.e., $\bar{F}_U(u) = e^{-\gamma u}$ for $u >0$.

Interestingly, this is a simple and efficient way to generate a couple of random variables with tail dependence.