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I'm wondering if Jensen's Inequality is transitive with regard to applying the same cocave-increasing transformation on two random variables, both of which have monotonically increasing density functions defined over a finite interval on the positive reals (NOTE: each variable can have a different interval, but both must be positive and finite). I.e., if E(X) < E(Y), and under a concave-increasing transformation f(x), E(f(X)) < E(X) and E(f(Y)) < E(Y), then can we conclude the E(f(X)) < E(f(Y))?

Thanks :)

Annika
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    Actually, Jensen's inequality, as stated in Wikipedia, says that if $f$ is a convex function and $X$ is a random variable, $$f(E[X]) \leq E[f(X)].$$ I believe your statement about $E[X] > E[f(X)]$ if $f$ is strictly concave is not true: for instance $X \equiv 0, f(x) = 1 - x^2$ – Pedro M. Aug 02 '13 at 16:41
  • I've edited my post to be as specific as possible to my problem. I'm focused on only monotonic, increasing probability density functions defined on the non-negative reals and transformations that are strictly convex-increasing. – Annika Aug 02 '13 at 17:32
  • The attached questionand Did's answer for the exponential distribution seems to imply that the above conjecture is correct. http://math.stackexchange.com/questions/412418/mean-of-an-increasing-function-over-exponential-distribution?rq=1 – Annika Aug 02 '13 at 17:37
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    @Annika Do you mean monotonically increasing cumulative distribution functions (instead of probability density functions)? – Pedro M. Aug 02 '13 at 18:11
  • @Annika I think you might have misunderstood my comment. When $X \equiv 0, f(x) = 1-x^2$, we see $E[f(X)] = 1 > E[X] = 0$ (so the statement is false). It is true, however, that $E[f(X)] \leq f(E[X]).$ – Pedro M. Aug 02 '13 at 18:11
  • @Pedro Milet - thanks for the clarification, I now see what you were referring to. In my revised question, I am referring to density functions that are monotonically increasing on [0, a] with a < infinity and monotonically increasing concave functions f(x). This is slightly different than your counterexample, which involved a point mass distribution with decreasing function. thanks for your insights into the more general statement....I'm not sure at this point if the more specific one will have a similar counteraxample. – Annika Aug 02 '13 at 18:19
  • There's no such thing as a probability density function that increases over the whole real line. Nor one that decreases over the whole real line. If $f$ is a probability density on the real line, then $\lim_{x\to\pm\infty} f(x) = 0$. Possibly you mean c.d.f. instead of density? – Michael Hardy Aug 02 '13 at 20:15
  • @Michael Hardy - my apologies, my reference to the reals did not mean [0, infinity), just that the variable had to be positive. These types of distributions would all be truncated at some maximum value and have a minimum of at least 0. I've edited my question to clarify this. – Annika Aug 02 '13 at 20:22
  • @MichaelHardy "If f is a probability density on the real line, then limx→±∞f(x)=0"... This is not true. – Did Aug 02 '13 at 20:28
  • @Did - perhaps Michael Hardy is referring to RV's with smooth denstities on infinite domains (e.g., the normal distribution). Surely such an RV on (-inf,+inf) cannot have a non-zero limit at infinity, otherwise, how would it integrate to one. Could you provide an example that demonstrates otherwise? – Annika Aug 02 '13 at 20:35
  • @Annika Smoothness is not the subject. You are making the mistake of believing that either the limit of $f$ is zero or the limit of $f$ is some $c\ne0$. But it may well happen that $f$ has no limit... Scary, eh? – Did Aug 02 '13 at 20:37
  • OK, not quite true as written: you can have spikes of height $1$ of increasing narrowness at an unbounded set of locations. In a sort of Cesaro-like way, these things approach $0$. And of course, you can have a measure-$0$ set of exceptional points, but we shouldn't care about those. At any rate, we can't have a density that always increases or always decreases. – Michael Hardy Aug 02 '13 at 20:40
  • If the limit exists, it is $0$; if it doesn't exist, then (I think) there's a sort of Cesaro-limit that is $0$. (Should I have some accents on "Cesaro"?) – Michael Hardy Aug 02 '13 at 20:42
  • @Did - that is interesting...I haven't run across one like that...I am assuming that, like Michael Hardy said, those features end up having measure 0? – Annika Aug 02 '13 at 20:44
  • @MichaelHardy Cesàro. (And of course $t^{-1}\int\limits_0^tf\to0$ when $t\to\pm\infty$ since $f$ is a density...) – Did Aug 02 '13 at 21:12
  • @Annika "those features end up having measure 0" Sorry but I do not know what that means (and MH did not say that, as far as I can tell). Since this is at best tangential to your question, let us leave it at that. – Did Aug 02 '13 at 21:13
  • @Annika : I wrote: "you can have spikes of height $1$ of increasing narrowness at an unbounded set of locations." This isn't about an exception that has measure $0$, but it's a case in which the limit of the density as $x\to\infty$ or $x\to-\infty$ doesn't exist at all. So in those cases where the limit exists, then it is $0$, but in cases like what I described here, it doesn't exist. – Michael Hardy Aug 02 '13 at 22:06

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Here is a counter-example: Let $X$ be a random variable taking values on $[0,a]$. Then clearly $E[X] \leq a$.

Now consider the function $f(x) = \ln(e^{a + 1} + x)$. $f$ is concave, and $f(X) \geq a + 1$, hence $$E[f(X)] \geq a + 1 > a \geq E[X]$$

Pedro M.
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  • Although that does not answer your actual question, which might be true or not. – Pedro M. Aug 02 '13 at 18:38
  • Thanks - I agree that strict concavity is not sufficient to ensure E[X] > E[g(X)], as you can form a function that ensures that min(g(X)) > E[X] – Annika Aug 02 '13 at 19:01