is this inequality valid for any $X$ and $Y$?
$begingroup$
Suppose $X$ and $Y$ are two continuous random variables with marginal
density $f(x)$ and $g(y)$. Is it true that
$$ E( ln f(X) ) geq E( ln g(X)) $$
this would be true iff $ln f(X) geq ln g(X)$ which is turn would be true as long as
$$ f(X) geq g(X) $$
but then we have nothing special about f and g, just that they are densities. I feel this result may be false, but I cant find a counterexample. Any ideas?
probability
asked Dec 17 '18 at 20:47
Mikey SpivakMikey Spivak
381215
$endgroup$
add a comment |
$begingroup$
Suppose $X$ and $Y$ are two continuous random variables with marginal
density $f(x)$ and $g(y)$. Is it true that
$$ E( ln f(X) ) geq E( ln g(X)) $$
this would be true iff $ln f(X) geq ln g(X)$ which is turn would be true as long as
$$ f(X) geq g(X) $$
but then we have nothing special about f and g, just that they are densities. I feel this result may be false, but I cant find a counterexample. Any ideas?
probability
asked Dec 17 '18 at 20:47
Mikey SpivakMikey Spivak
381215
$endgroup$
1
$begingroup$
I mean, if you just switch the names f,g you have inequality in the other direction, so your only concern is then equality for all f,g.
$endgroup$
– T. Fo
Dec 17 '18 at 20:52
1
$begingroup$
As mentioned, since the statement is symmetric, you are actually asking if $$mathbb{E}[ln f(X)] = mathbb{E}[ln g(Y)]$$ for every two continuous r.v.'s $X,Y$ with densities$f,g$ respectively. It's simple to check with counterexamples: $X$ uniform on $[0,1]$ gives $mathbb{E}[ln f(X)] = 0$, $Y$ uniform on $[0,2]$ gives $mathbb{E}[ln g(Y)] = -ln 2$. (Note that for these two, $f,g$ are constant.)
$endgroup$
– Clement C.
Dec 17 '18 at 21:28
add a comment |
$begingroup$
Suppose $X$ and $Y$ are two continuous random variables with marginal
density $f(x)$ and $g(y)$. Is it true that
$$ E( ln f(X) ) geq E( ln g(X)) $$
this would be true iff $ln f(X) geq ln g(X)$ which is turn would be true as long as
$$ f(X) geq g(X) $$
but then we have nothing special about f and g, just that they are densities. I feel this result may be false, but I cant find a counterexample. Any ideas?
probability
asked Dec 17 '18 at 20:47
Mikey SpivakMikey Spivak
381215
$endgroup$
Suppose $X$ and $Y$ are two continuous random variables with marginal
density $f(x)$ and $g(y)$. Is it true that
$$ E( ln f(X) ) geq E( ln g(X)) $$
this would be true iff $ln f(X) geq ln g(X)$ which is turn would be true as long as
$$ f(X) geq g(X) $$
but then we have nothing special about f and g, just that they are densities. I feel this result may be false, but I cant find a counterexample. Any ideas?
probability
probability
asked Dec 17 '18 at 20:47
Mikey SpivakMikey Spivak
381215
asked Dec 17 '18 at 20:47
Mikey SpivakMikey Spivak
381215
asked Dec 17 '18 at 20:47
Mikey SpivakMikey Spivak
381215
asked Dec 17 '18 at 20:47
Mikey SpivakMikey Spivak
381215
asked Dec 17 '18 at 20:47
Mikey SpivakMikey Spivak
381215
381215
1
$begingroup$
I mean, if you just switch the names f,g you have inequality in the other direction, so your only concern is then equality for all f,g.
$endgroup$
– T. Fo
Dec 17 '18 at 20:52
1
$begingroup$
As mentioned, since the statement is symmetric, you are actually asking if $$mathbb{E}[ln f(X)] = mathbb{E}[ln g(Y)]$$ for every two continuous r.v.'s $X,Y$ with densities$f,g$ respectively. It's simple to check with counterexamples: $X$ uniform on $[0,1]$ gives $mathbb{E}[ln f(X)] = 0$, $Y$ uniform on $[0,2]$ gives $mathbb{E}[ln g(Y)] = -ln 2$. (Note that for these two, $f,g$ are constant.)
$endgroup$
– Clement C.
Dec 17 '18 at 21:28
add a comment |
1
$begingroup$
I mean, if you just switch the names f,g you have inequality in the other direction, so your only concern is then equality for all f,g.
$endgroup$
– T. Fo
Dec 17 '18 at 20:52
1
$begingroup$
As mentioned, since the statement is symmetric, you are actually asking if $$mathbb{E}[ln f(X)] = mathbb{E}[ln g(Y)]$$ for every two continuous r.v.'s $X,Y$ with densities$f,g$ respectively. It's simple to check with counterexamples: $X$ uniform on $[0,1]$ gives $mathbb{E}[ln f(X)] = 0$, $Y$ uniform on $[0,2]$ gives $mathbb{E}[ln g(Y)] = -ln 2$. (Note that for these two, $f,g$ are constant.)
$endgroup$
– Clement C.
Dec 17 '18 at 21:28
1
1
$begingroup$
I mean, if you just switch the names f,g you have inequality in the other direction, so your only concern is then equality for all f,g.
$endgroup$
– T. Fo
Dec 17 '18 at 20:52
$begingroup$
I mean, if you just switch the names f,g you have inequality in the other direction, so your only concern is then equality for all f,g.
$endgroup$
– T. Fo
Dec 17 '18 at 20:52
1
1
$begingroup$
As mentioned, since the statement is symmetric, you are actually asking if $$mathbb{E}[ln f(X)] = mathbb{E}[ln g(Y)]$$ for every two continuous r.v.'s $X,Y$ with densities$f,g$ respectively. It's simple to check with counterexamples: $X$ uniform on $[0,1]$ gives $mathbb{E}[ln f(X)] = 0$, $Y$ uniform on $[0,2]$ gives $mathbb{E}[ln g(Y)] = -ln 2$. (Note that for these two, $f,g$ are constant.)
$endgroup$
– Clement C.
Dec 17 '18 at 21:28
$begingroup$
As mentioned, since the statement is symmetric, you are actually asking if $$mathbb{E}[ln f(X)] = mathbb{E}[ln g(Y)]$$ for every two continuous r.v.'s $X,Y$ with densities$f,g$ respectively. It's simple to check with counterexamples: $X$ uniform on $[0,1]$ gives $mathbb{E}[ln f(X)] = 0$, $Y$ uniform on $[0,2]$ gives $mathbb{E}[ln g(Y)] = -ln 2$. (Note that for these two, $f,g$ are constant.)
$endgroup$
– Clement C.
Dec 17 '18 at 21:28
add a comment |
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1
$begingroup$
I mean, if you just switch the names f,g you have inequality in the other direction, so your only concern is then equality for all f,g.
$endgroup$
– T. Fo
Dec 17 '18 at 20:52
1
$begingroup$
As mentioned, since the statement is symmetric, you are actually asking if $$mathbb{E}[ln f(X)] = mathbb{E}[ln g(Y)]$$ for every two continuous r.v.'s $X,Y$ with densities$f,g$ respectively. It's simple to check with counterexamples: $X$ uniform on $[0,1]$ gives $mathbb{E}[ln f(X)] = 0$, $Y$ uniform on $[0,2]$ gives $mathbb{E}[ln g(Y)] = -ln 2$. (Note that for these two, $f,g$ are constant.)
$endgroup$
– Clement C.
Dec 17 '18 at 21:28