Difficult probability question in game theory that involves two random variables
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I am currently studying probabilities while doing some game theory, and I'm having trouble solving a particular question that I came up with.
Suppose you have two random variables, $sigma_1$ and $sigma_2$, where $sigma_1$ and $sigma_2$ are independently distributed. Suppose further that $sigma_1$ is uniformly distributed in an interval $[-phi, phi]$ and $sigma_2$ is uniformly distributed in an interval $[0,bar{sigma}_2]$ (not too important here).
Now, let there be some constants $a,b,c$.
How would we calculate this probability?
$p(2b-sigma_1 +2c geq 2sigma_2)$ and $p(b-sigma_1 geq a)$
Here's what I have so far. Let $mathbb{1}$ denote the indicator function and let $F(sigma_1)$ and $G(sigma_2)$ represent the distribution functions:
$int_{sigma_1} int_{sigma_2} mathbb{1} [sigma_1 leq min{2b+2c-2sigma_2, b - a}] dG(sigma_2) dF(sigma_1)$
Assuming that $bar{sigma}_2$ is large enough ($bar{sigma}_2 > b-a$), can I write this as:
$int_{sigma_1} left[int_{b-a}^{bar{sigma}_2} mathbb{1}[sigma_1 leq b - a] dG(sigma_2) + int_{0}^{b-a} mathbb{1}[ sigma_1 leq 2b+2c-2sigma_2] dG(sigma_2)right] dF(sigma_1)$
Is this correct? Or can anyone help clarify how to simplify this?
Thank you
probability-theory game-theory
asked Nov 20 at 9:11
Steve
12
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0
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I am currently studying probabilities while doing some game theory, and I'm having trouble solving a particular question that I came up with.
Suppose you have two random variables, $sigma_1$ and $sigma_2$, where $sigma_1$ and $sigma_2$ are independently distributed. Suppose further that $sigma_1$ is uniformly distributed in an interval $[-phi, phi]$ and $sigma_2$ is uniformly distributed in an interval $[0,bar{sigma}_2]$ (not too important here).
Now, let there be some constants $a,b,c$.
How would we calculate this probability?
$p(2b-sigma_1 +2c geq 2sigma_2)$ and $p(b-sigma_1 geq a)$
Here's what I have so far. Let $mathbb{1}$ denote the indicator function and let $F(sigma_1)$ and $G(sigma_2)$ represent the distribution functions:
$int_{sigma_1} int_{sigma_2} mathbb{1} [sigma_1 leq min{2b+2c-2sigma_2, b - a}] dG(sigma_2) dF(sigma_1)$
Assuming that $bar{sigma}_2$ is large enough ($bar{sigma}_2 > b-a$), can I write this as:
$int_{sigma_1} left[int_{b-a}^{bar{sigma}_2} mathbb{1}[sigma_1 leq b - a] dG(sigma_2) + int_{0}^{b-a} mathbb{1}[ sigma_1 leq 2b+2c-2sigma_2] dG(sigma_2)right] dF(sigma_1)$
Is this correct? Or can anyone help clarify how to simplify this?
Thank you
probability-theory game-theory
asked Nov 20 at 9:11
Steve
12
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am currently studying probabilities while doing some game theory, and I'm having trouble solving a particular question that I came up with.
Suppose you have two random variables, $sigma_1$ and $sigma_2$, where $sigma_1$ and $sigma_2$ are independently distributed. Suppose further that $sigma_1$ is uniformly distributed in an interval $[-phi, phi]$ and $sigma_2$ is uniformly distributed in an interval $[0,bar{sigma}_2]$ (not too important here).
Now, let there be some constants $a,b,c$.
How would we calculate this probability?
$p(2b-sigma_1 +2c geq 2sigma_2)$ and $p(b-sigma_1 geq a)$
Here's what I have so far. Let $mathbb{1}$ denote the indicator function and let $F(sigma_1)$ and $G(sigma_2)$ represent the distribution functions:
$int_{sigma_1} int_{sigma_2} mathbb{1} [sigma_1 leq min{2b+2c-2sigma_2, b - a}] dG(sigma_2) dF(sigma_1)$
Assuming that $bar{sigma}_2$ is large enough ($bar{sigma}_2 > b-a$), can I write this as:
$int_{sigma_1} left[int_{b-a}^{bar{sigma}_2} mathbb{1}[sigma_1 leq b - a] dG(sigma_2) + int_{0}^{b-a} mathbb{1}[ sigma_1 leq 2b+2c-2sigma_2] dG(sigma_2)right] dF(sigma_1)$
Is this correct? Or can anyone help clarify how to simplify this?
Thank you
probability-theory game-theory
asked Nov 20 at 9:11
Steve
12
I am currently studying probabilities while doing some game theory, and I'm having trouble solving a particular question that I came up with.
Suppose you have two random variables, $sigma_1$ and $sigma_2$, where $sigma_1$ and $sigma_2$ are independently distributed. Suppose further that $sigma_1$ is uniformly distributed in an interval $[-phi, phi]$ and $sigma_2$ is uniformly distributed in an interval $[0,bar{sigma}_2]$ (not too important here).
Now, let there be some constants $a,b,c$.
How would we calculate this probability?
$p(2b-sigma_1 +2c geq 2sigma_2)$ and $p(b-sigma_1 geq a)$
Here's what I have so far. Let $mathbb{1}$ denote the indicator function and let $F(sigma_1)$ and $G(sigma_2)$ represent the distribution functions:
$int_{sigma_1} int_{sigma_2} mathbb{1} [sigma_1 leq min{2b+2c-2sigma_2, b - a}] dG(sigma_2) dF(sigma_1)$
Assuming that $bar{sigma}_2$ is large enough ($bar{sigma}_2 > b-a$), can I write this as:
$int_{sigma_1} left[int_{b-a}^{bar{sigma}_2} mathbb{1}[sigma_1 leq b - a] dG(sigma_2) + int_{0}^{b-a} mathbb{1}[ sigma_1 leq 2b+2c-2sigma_2] dG(sigma_2)right] dF(sigma_1)$
Is this correct? Or can anyone help clarify how to simplify this?
Thank you
probability-theory game-theory
probability-theory game-theory
asked Nov 20 at 9:11
Steve
12
asked Nov 20 at 9:11
Steve
12
edited Nov 20 at 9:19
asked Nov 20 at 9:11
Steve
12
asked Nov 20 at 9:11
Steve
12
asked Nov 20 at 9:11
Steve
12
12
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