Mixing
Applying mean value theorem 1 to expectation.
$begingroup$
I would like to prove that for an arbitrary real random variable $X$ defined on some continuous and closed interval $S$, if functions $h(x)$ and $g(x)$ are bounded, continuous and non-zero over $S$, we can find some $x_0 in S$ such that
$$
frac{mathbb{E}[h(X)]}{mathbb{E}[g(X)]} = frac{h(x_0)}{g(x_0)}
$$
$X$ can be either continuous or discrete.
My proof is
$$
frac{mathbb{E}[h(X)]}{mathbb{E}[g(X)]} = frac{int_S h(x) d F_X(x)}{int_S g(x) d F_X(x)} = frac{int_S frac{h(x)}{g(x)} g(x) d F_X(x)}{int_S g(x) d F_X(x)} = frac{frac{h(x_0)}{g(x_0)} int_S g(x) d F_X(x)}{int_S g(x) d F_X(x)} = frac{h(x_0)}{g(x_0)}
$$
because $h(x)/g(x)$ is continuous on $S$. Is there anything I am missing? Not an expert in probability, so I am wondering if there is any formal theorem for this result.
probability integration
edited Jan 19 at 23:49
David G. Stork
11k41432
asked Jan 19 at 23:39
LiangLiang
907
$endgroup$
add a comment |
$begingroup$
I would like to prove that for an arbitrary real random variable $X$ defined on some continuous and closed interval $S$, if functions $h(x)$ and $g(x)$ are bounded, continuous and non-zero over $S$, we can find some $x_0 in S$ such that
$$
frac{mathbb{E}[h(X)]}{mathbb{E}[g(X)]} = frac{h(x_0)}{g(x_0)}
$$
$X$ can be either continuous or discrete.
My proof is
$$
frac{mathbb{E}[h(X)]}{mathbb{E}[g(X)]} = frac{int_S h(x) d F_X(x)}{int_S g(x) d F_X(x)} = frac{int_S frac{h(x)}{g(x)} g(x) d F_X(x)}{int_S g(x) d F_X(x)} = frac{frac{h(x_0)}{g(x_0)} int_S g(x) d F_X(x)}{int_S g(x) d F_X(x)} = frac{h(x_0)}{g(x_0)}
$$
because $h(x)/g(x)$ is continuous on $S$. Is there anything I am missing? Not an expert in probability, so I am wondering if there is any formal theorem for this result.
probability integration
edited Jan 19 at 23:49
David G. Stork
11k41432
asked Jan 19 at 23:39
LiangLiang
907
$endgroup$
$begingroup$
This mean value theorem only holds if h/g does not change sign on S
$endgroup$
– Paul Cottalorda
Jan 20 at 1:00
$begingroup$
I think that your proof is correct. $h/g$ is continuous on $S$ so the (first) MVT for integrals holds. BTW, $h/g$ does not change sign (because both $f$ and $g$ preserve the sign by continuity ) but I do not think that is relevant.
$endgroup$
– GReyes
Jan 20 at 8:04
add a comment |
$begingroup$
I would like to prove that for an arbitrary real random variable $X$ defined on some continuous and closed interval $S$, if functions $h(x)$ and $g(x)$ are bounded, continuous and non-zero over $S$, we can find some $x_0 in S$ such that
$$
frac{mathbb{E}[h(X)]}{mathbb{E}[g(X)]} = frac{h(x_0)}{g(x_0)}
$$
$X$ can be either continuous or discrete.
My proof is
$$
frac{mathbb{E}[h(X)]}{mathbb{E}[g(X)]} = frac{int_S h(x) d F_X(x)}{int_S g(x) d F_X(x)} = frac{int_S frac{h(x)}{g(x)} g(x) d F_X(x)}{int_S g(x) d F_X(x)} = frac{frac{h(x_0)}{g(x_0)} int_S g(x) d F_X(x)}{int_S g(x) d F_X(x)} = frac{h(x_0)}{g(x_0)}
$$
because $h(x)/g(x)$ is continuous on $S$. Is there anything I am missing? Not an expert in probability, so I am wondering if there is any formal theorem for this result.
probability integration
edited Jan 19 at 23:49
David G. Stork
11k41432
asked Jan 19 at 23:39
LiangLiang
907
$endgroup$
I would like to prove that for an arbitrary real random variable $X$ defined on some continuous and closed interval $S$, if functions $h(x)$ and $g(x)$ are bounded, continuous and non-zero over $S$, we can find some $x_0 in S$ such that
$$
frac{mathbb{E}[h(X)]}{mathbb{E}[g(X)]} = frac{h(x_0)}{g(x_0)}
$$
$X$ can be either continuous or discrete.
My proof is
$$
frac{mathbb{E}[h(X)]}{mathbb{E}[g(X)]} = frac{int_S h(x) d F_X(x)}{int_S g(x) d F_X(x)} = frac{int_S frac{h(x)}{g(x)} g(x) d F_X(x)}{int_S g(x) d F_X(x)} = frac{frac{h(x_0)}{g(x_0)} int_S g(x) d F_X(x)}{int_S g(x) d F_X(x)} = frac{h(x_0)}{g(x_0)}
$$
because $h(x)/g(x)$ is continuous on $S$. Is there anything I am missing? Not an expert in probability, so I am wondering if there is any formal theorem for this result.
probability integration
probability integration
edited Jan 19 at 23:49
David G. Stork
11k41432
asked Jan 19 at 23:39
LiangLiang
907
edited Jan 19 at 23:49
David G. Stork
11k41432
asked Jan 19 at 23:39
LiangLiang
907
edited Jan 19 at 23:49
David G. Stork
11k41432
edited Jan 19 at 23:49
David G. Stork
11k41432
edited Jan 19 at 23:49
David G. Stork
11k41432
11k41432
asked Jan 19 at 23:39
LiangLiang
907
asked Jan 19 at 23:39
LiangLiang
907
asked Jan 19 at 23:39
LiangLiang
907
907
$begingroup$
This mean value theorem only holds if h/g does not change sign on S
$endgroup$
– Paul Cottalorda
Jan 20 at 1:00
$begingroup$
I think that your proof is correct. $h/g$ is continuous on $S$ so the (first) MVT for integrals holds. BTW, $h/g$ does not change sign (because both $f$ and $g$ preserve the sign by continuity ) but I do not think that is relevant.
$endgroup$
– GReyes
Jan 20 at 8:04
add a comment |
$begingroup$
This mean value theorem only holds if h/g does not change sign on S
$endgroup$
– Paul Cottalorda
Jan 20 at 1:00
$begingroup$
I think that your proof is correct. $h/g$ is continuous on $S$ so the (first) MVT for integrals holds. BTW, $h/g$ does not change sign (because both $f$ and $g$ preserve the sign by continuity ) but I do not think that is relevant.
$endgroup$
– GReyes
Jan 20 at 8:04
$begingroup$
This mean value theorem only holds if h/g does not change sign on S
$endgroup$
– Paul Cottalorda
Jan 20 at 1:00
$begingroup$
This mean value theorem only holds if h/g does not change sign on S
$endgroup$
– Paul Cottalorda
Jan 20 at 1:00
$begingroup$
I think that your proof is correct. $h/g$ is continuous on $S$ so the (first) MVT for integrals holds. BTW, $h/g$ does not change sign (because both $f$ and $g$ preserve the sign by continuity ) but I do not think that is relevant.
$endgroup$
– GReyes
Jan 20 at 8:04
$begingroup$
I think that your proof is correct. $h/g$ is continuous on $S$ so the (first) MVT for integrals holds. BTW, $h/g$ does not change sign (because both $f$ and $g$ preserve the sign by continuity ) but I do not think that is relevant.
$endgroup$
– GReyes
Jan 20 at 8:04
add a comment |
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$begingroup$
This mean value theorem only holds if h/g does not change sign on S
$endgroup$
– Paul Cottalorda
Jan 20 at 1:00
$begingroup$
I think that your proof is correct. $h/g$ is continuous on $S$ so the (first) MVT for integrals holds. BTW, $h/g$ does not change sign (because both $f$ and $g$ preserve the sign by continuity ) but I do not think that is relevant.
$endgroup$
– GReyes
Jan 20 at 8:04