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Davide Giraudo
Davide Giraudo
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Assume that the random variables $X$ and $Y$ are defined on the probability space $(\Omega,\mathcal F,\mu)$. Let $\Delta:=\{(x,y)\in\Bbb R^2,x\lt y\}$. We have by independence $$ E\left[e^{it\max(X,Y)}\right]=\int_{\Bbb R^2}e^{it\max(x,y)}\mathrm d\mu_X\otimes\mu_Y(x,y). $$ Splitting over $\Delta$ and its complement, and denoting $F$ the common cumulative distribution function of $X$ and $Y$, we thus get $$E\left[e^{it\max(X,Y)}\right]=2E\left[F(X)e^{itX}\right]+\int_{\Bbb R}\mu(X=x)e^{itx}\mathrm d\mu(x).$$$$E\left[e^{it\max(X,Y)}\right]=2E\left[F(X)e^{itX}\right]-\int_{\Bbb R}\mu(X=x)e^{itx}\mathrm d\mu_X(x).$$

Some remarks:

  1. This gives an explicit formula in terms of the common distribution function.
  2. If $\mu(X=x)=0$ for all $x$ (for example when $X$ has a density), then the formula is simpler.
  3. This can be extended to $\max(X_1,\dots,X_d)$.
  4. We get an analogous formula for $\min$ instead of $\max$.

Assume that the random variables $X$ and $Y$ are defined on the probability space $(\Omega,\mathcal F,\mu)$. Let $\Delta:=\{(x,y)\in\Bbb R^2,x\lt y\}$. We have by independence $$ E\left[e^{it\max(X,Y)}\right]=\int_{\Bbb R^2}e^{it\max(x,y)}\mathrm d\mu_X\otimes\mu_Y(x,y). $$ Splitting over $\Delta$ and its complement, and denoting $F$ the common cumulative distribution function of $X$ and $Y$, we thus get $$E\left[e^{it\max(X,Y)}\right]=2E\left[F(X)e^{itX}\right]+\int_{\Bbb R}\mu(X=x)e^{itx}\mathrm d\mu(x).$$

Some remarks:

  1. This gives an explicit formula in terms of the common distribution function.
  2. If $\mu(X=x)=0$ for all $x$ (for example when $X$ has a density), then the formula is simpler.
  3. This can be extended to $\max(X_1,\dots,X_d)$.
  4. We get an analogous formula for $\min$ instead of $\max$.

Assume that the random variables $X$ and $Y$ are defined on the probability space $(\Omega,\mathcal F,\mu)$. Let $\Delta:=\{(x,y)\in\Bbb R^2,x\lt y\}$. We have by independence $$ E\left[e^{it\max(X,Y)}\right]=\int_{\Bbb R^2}e^{it\max(x,y)}\mathrm d\mu_X\otimes\mu_Y(x,y). $$ Splitting over $\Delta$ and its complement, and denoting $F$ the common cumulative distribution function of $X$ and $Y$, we thus get $$E\left[e^{it\max(X,Y)}\right]=2E\left[F(X)e^{itX}\right]-\int_{\Bbb R}\mu(X=x)e^{itx}\mathrm d\mu_X(x).$$

Some remarks:

  1. This gives an explicit formula in terms of the common distribution function.
  2. If $\mu(X=x)=0$ for all $x$ (for example when $X$ has a density), then the formula is simpler.
  3. This can be extended to $\max(X_1,\dots,X_d)$.
  4. We get an analogous formula for $\min$ instead of $\max$.
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Davide Giraudo
Davide Giraudo
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Assume that the random variables $X$ and $Y$ are defined on the probability space $(\Omega,\mathcal F,\mu)$. Let $\Delta:=((x,y)\in\Bbb R^2,x\lt y)$$\Delta:=\{(x,y)\in\Bbb R^2,x\lt y\}$. We have by independence $$ E\left\[e^{it\max(X,Y)}\right\]=\int_{\Bbb R^2}e^{it\max(x,y)}\mathrm d\mu_X\otimes\mu_Y(x,y). $$$$ E\left[e^{it\max(X,Y)}\right]=\int_{\Bbb R^2}e^{it\max(x,y)}\mathrm d\mu_X\otimes\mu_Y(x,y). $$ Splitting over $\Delta$ and its complement, and denoting $F$ the common cumulative distribution function of $X$ and $Y$, we thus get $$E\left\[e^{it\max(X,Y)}\right\]=2E\left[F(X)e^{itX}\right]+\int_{\Bbb R}\mu(X=x)e^{itx}\mathrm d\mu(x).$$$$E\left[e^{it\max(X,Y)}\right]=2E\left[F(X)e^{itX}\right]+\int_{\Bbb R}\mu(X=x)e^{itx}\mathrm d\mu(x).$$

Some remarks:

  1. This involves only information on $X$gives an explicit formula in terms of the common distribution function.
  2. If $\mu(X=x)=0$ for all $x$ (for example when $X$ has a density), then the formula is simpler.
  3. This can be extended to $\max(X_1,\dots,X_d)$.
  4. We get an analogous formula for $\min$ instead of $\max$.

Assume that the random variables $X$ and $Y$ are defined on the probability space $(\Omega,\mathcal F,\mu)$. Let $\Delta:=((x,y)\in\Bbb R^2,x\lt y)$. We have by independence $$ E\left\[e^{it\max(X,Y)}\right\]=\int_{\Bbb R^2}e^{it\max(x,y)}\mathrm d\mu_X\otimes\mu_Y(x,y). $$ Splitting over $\Delta$ and its complement, and denoting $F$ the common cumulative distribution function of $X$ and $Y$, we thus get $$E\left\[e^{it\max(X,Y)}\right\]=2E\left[F(X)e^{itX}\right]+\int_{\Bbb R}\mu(X=x)e^{itx}\mathrm d\mu(x).$$

Some remarks:

  1. This involves only information on $X$.
  2. If $\mu(X=x)=0$ for all $x$ (for example when $X$ has a density), then the formula is simpler.
  3. This can be extended to $\max(X_1,\dots,X_d)$.
  4. We get an analogous formula for $\min$ instead of $\max$.

Assume that the random variables $X$ and $Y$ are defined on the probability space $(\Omega,\mathcal F,\mu)$. Let $\Delta:=\{(x,y)\in\Bbb R^2,x\lt y\}$. We have by independence $$ E\left[e^{it\max(X,Y)}\right]=\int_{\Bbb R^2}e^{it\max(x,y)}\mathrm d\mu_X\otimes\mu_Y(x,y). $$ Splitting over $\Delta$ and its complement, and denoting $F$ the common cumulative distribution function of $X$ and $Y$, we thus get $$E\left[e^{it\max(X,Y)}\right]=2E\left[F(X)e^{itX}\right]+\int_{\Bbb R}\mu(X=x)e^{itx}\mathrm d\mu(x).$$

Some remarks:

  1. This gives an explicit formula in terms of the common distribution function.
  2. If $\mu(X=x)=0$ for all $x$ (for example when $X$ has a density), then the formula is simpler.
  3. This can be extended to $\max(X_1,\dots,X_d)$.
  4. We get an analogous formula for $\min$ instead of $\max$.
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Davide Giraudo
Davide Giraudo
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Assume that the random variables $X$ and $Y$ are defined on the probability space $(\Omega,\mathcal F,\mu)$. Let $\Delta:=((x,y)\in\Bbb R^2,x\lt y)$. We have by independence $$ E\left\[e^{it\max(X,Y)}\right\]=\int_{\Bbb R^2}e^{it\max(x,y)}\mathrm d\mu_X\otimes\mu_Y(x,y). $$ Splitting over $\Delta$ and its complement, and denoting $F$ the common cumulative distribution function of $X$ and $Y$, we thus get $$E\left\[e^{it\max(X,Y)}\right\]=2E\left[F(X)e^{itX}\right]+\int_{\Bbb R}\mu(X=x)e^{itx}\mathrm d\mu(x).$$

Some remarks:

  1. This involves only information on $X$.
  2. If $\mu(X=x)=0$ for all $x$ (for example when $X$ has a density), then the formula is simpler.
  3. This can be extended to $\max(X_1,\dots,X_d)$.
  4. We get an analogous formula for $\min$ instead of $\max$.

Assume that the random variables $X$ and $Y$ are defined on the probability space $(\Omega,\mathcal F,\mu)$. Let $\Delta:=((x,y)\in\Bbb R^2,x\lt y)$. We have by independence $$ E\left\[e^{it\max(X,Y)}\right\]=\int_{\Bbb R^2}e^{it\max(x,y)}\mathrm d\mu_X\otimes\mu_Y(x,y). $$ Splitting over $\Delta$ and its complement, and denoting $F$ the common cumulative distribution function of $X$ and $Y$, we thus get $$E\left\[e^{it\max(X,Y)}\right\]=2E\left[F(X)e^{itX}\right]+\int_{\Bbb R}\mu(X=x)e^{itx}\mathrm d\mu(x).$$

  1. This involves only information on $X$.
  2. If $\mu(X=x)=0$ for all $x$ (for example when $X$ has a density), then the formula is simpler.
  3. This can be extended to $\max(X_1,\dots,X_d)$.

Assume that the random variables $X$ and $Y$ are defined on the probability space $(\Omega,\mathcal F,\mu)$. Let $\Delta:=((x,y)\in\Bbb R^2,x\lt y)$. We have by independence $$ E\left\[e^{it\max(X,Y)}\right\]=\int_{\Bbb R^2}e^{it\max(x,y)}\mathrm d\mu_X\otimes\mu_Y(x,y). $$ Splitting over $\Delta$ and its complement, and denoting $F$ the common cumulative distribution function of $X$ and $Y$, we thus get $$E\left\[e^{it\max(X,Y)}\right\]=2E\left[F(X)e^{itX}\right]+\int_{\Bbb R}\mu(X=x)e^{itx}\mathrm d\mu(x).$$

Some remarks:

  1. This involves only information on $X$.
  2. If $\mu(X=x)=0$ for all $x$ (for example when $X$ has a density), then the formula is simpler.
  3. This can be extended to $\max(X_1,\dots,X_d)$.
  4. We get an analogous formula for $\min$ instead of $\max$.
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Davide Giraudo
Davide Giraudo
  • 4k
  • 2
  • 30
  • 43