relation between $P(mathbf{X}geq t)$ and $P(Phi(mathbf{X}) geq Phi(t))$
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The relation between $P(mathbf{X}geq t)$ and $P(Phi(mathbf{X}) geq Phi(t))$ for any non-decreasing non-negative function $Phi$, non-negative $t$ and for positive random variables $mathbf{X}$ is such that
$$
P(Phi(mathbf{X}) geq Phi(t)) geq P(mathbf{X}geq t)
$$
For functions like $Phi(x) = x^2$, it can be easily imagine such an inequality as the even space increases. For example if $x in (0,2)$ then $x^2 in (0,4)$. But what happens in case of other functions like $Phi(x) = sqrt{x}$ ?. Here event space decreases, if $x in (0,4)$, then $sqrt{x} in (0,2)$ only. In such situation how one can show that above inequality holds ?
probability-theory inequality
asked Nov 22 at 14:34
Shew
553413
add a comment |
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The relation between $P(mathbf{X}geq t)$ and $P(Phi(mathbf{X}) geq Phi(t))$ for any non-decreasing non-negative function $Phi$, non-negative $t$ and for positive random variables $mathbf{X}$ is such that
$$
P(Phi(mathbf{X}) geq Phi(t)) geq P(mathbf{X}geq t)
$$
For functions like $Phi(x) = x^2$, it can be easily imagine such an inequality as the even space increases. For example if $x in (0,2)$ then $x^2 in (0,4)$. But what happens in case of other functions like $Phi(x) = sqrt{x}$ ?. Here event space decreases, if $x in (0,4)$, then $sqrt{x} in (0,2)$ only. In such situation how one can show that above inequality holds ?
probability-theory inequality
asked Nov 22 at 14:34
Shew
553413
I learned probality theory long ago, so I may be wrong. But if $mathbf{X}geq t$ then since $Phi$ is non-decreasing, we have $Phi(mathbf{X})geq Phi(t)$ which implies the required inequality, right?
– Alex Ravsky
Nov 27 at 4:23
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The relation between $P(mathbf{X}geq t)$ and $P(Phi(mathbf{X}) geq Phi(t))$ for any non-decreasing non-negative function $Phi$, non-negative $t$ and for positive random variables $mathbf{X}$ is such that
$$
P(Phi(mathbf{X}) geq Phi(t)) geq P(mathbf{X}geq t)
$$
For functions like $Phi(x) = x^2$, it can be easily imagine such an inequality as the even space increases. For example if $x in (0,2)$ then $x^2 in (0,4)$. But what happens in case of other functions like $Phi(x) = sqrt{x}$ ?. Here event space decreases, if $x in (0,4)$, then $sqrt{x} in (0,2)$ only. In such situation how one can show that above inequality holds ?
probability-theory inequality
asked Nov 22 at 14:34
Shew
553413
The relation between $P(mathbf{X}geq t)$ and $P(Phi(mathbf{X}) geq Phi(t))$ for any non-decreasing non-negative function $Phi$, non-negative $t$ and for positive random variables $mathbf{X}$ is such that
$$
P(Phi(mathbf{X}) geq Phi(t)) geq P(mathbf{X}geq t)
$$
For functions like $Phi(x) = x^2$, it can be easily imagine such an inequality as the even space increases. For example if $x in (0,2)$ then $x^2 in (0,4)$. But what happens in case of other functions like $Phi(x) = sqrt{x}$ ?. Here event space decreases, if $x in (0,4)$, then $sqrt{x} in (0,2)$ only. In such situation how one can show that above inequality holds ?
probability-theory inequality
probability-theory inequality
asked Nov 22 at 14:34
Shew
553413
asked Nov 22 at 14:34
Shew
553413
asked Nov 22 at 14:34
Shew
553413
asked Nov 22 at 14:34
Shew
553413
asked Nov 22 at 14:34
Shew
553413
553413
I learned probality theory long ago, so I may be wrong. But if $mathbf{X}geq t$ then since $Phi$ is non-decreasing, we have $Phi(mathbf{X})geq Phi(t)$ which implies the required inequality, right?
– Alex Ravsky
Nov 27 at 4:23
add a comment |
I learned probality theory long ago, so I may be wrong. But if $mathbf{X}geq t$ then since $Phi$ is non-decreasing, we have $Phi(mathbf{X})geq Phi(t)$ which implies the required inequality, right?
– Alex Ravsky
Nov 27 at 4:23
I learned probality theory long ago, so I may be wrong. But if $mathbf{X}geq t$ then since $Phi$ is non-decreasing, we have $Phi(mathbf{X})geq Phi(t)$ which implies the required inequality, right?
– Alex Ravsky
Nov 27 at 4:23
I learned probality theory long ago, so I may be wrong. But if $mathbf{X}geq t$ then since $Phi$ is non-decreasing, we have $Phi(mathbf{X})geq Phi(t)$ which implies the required inequality, right?
– Alex Ravsky
Nov 27 at 4:23
add a comment |
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I learned probality theory long ago, so I may be wrong. But if $mathbf{X}geq t$ then since $Phi$ is non-decreasing, we have $Phi(mathbf{X})geq Phi(t)$ which implies the required inequality, right?
– Alex Ravsky
Nov 27 at 4:23