2 added 1 character in body
source | link

Others have already explained that the answer to the question is NO, except trivial cases. Below we give an approach to finding $\DeclareMathOperator{\E}{\mathbb{E}} \E \frac1{X}$ when $X>0$ with probability one, and the moment generating function $M_X(t) = \E e^{tX}$ do exist. An An application of this method (and a genaralizationgeneralization) is given in Expected value of $1/x$ when $x$ follows a Beta distribution, we will here also give a simpler example.

First, note that $\int_0^\infty e^{-t x}\; dt = \frac1{x}$ (simple calculuacalculus exercise). Then, write $$ \E \left(\frac1{X}\right) = \int_0^\infty x^{-1} f(x)\; dx = \int_0^\infty \left( \int_0^\infty e^{-tx}\; dt \right) f(x)\; dx =\\ \int_0^\infty \left( \int_0^\infty e^{-tx} f(x) \; dx \right) \; dt = \int_0^\infty M_X(-t) \; dt $$ A simple application: Let $X$ have the exponential distribution with rate 1, that is, with density $e^{-x}, x>0$ and moment generating function $M_X(t)=\frac1{1-t}, t<1$. Then $\int_0^\infty M_X(-t)\; dt = \int_0^\infty \frac1{1+t} \; dt= \ln(1+t) \bigg\rvert_0^\infty = \infty$, so definitely do not converge, and is very different from $\frac1{\E X}=\frac11=1$.

Others have already explained that the answer to the question is NO, except trivial cases. Below we give an approach to finding $\DeclareMathOperator{\E}{\mathbb{E}} \E \frac1{X}$ when $X>0$ with probability one, and the moment generating function $M_X(t) = \E e^{tX}$ do exist. An application of this method (and a genaralization) is given in Expected value of $1/x$ when $x$ follows a Beta distribution, we will here also give a simpler example.

First, note that $\int_0^\infty e^{-t x}\; dt = \frac1{x}$ (simple calculua exercise). Then, write $$ \E \left(\frac1{X}\right) = \int_0^\infty x^{-1} f(x)\; dx = \int_0^\infty \left( \int_0^\infty e^{-tx}\; dt \right) f(x)\; dx =\\ \int_0^\infty \left( \int_0^\infty e^{-tx} f(x) \; dx \right) \; dt = \int_0^\infty M_X(-t) \; dt $$ A simple application: Let $X$ have the exponential distribution with rate 1, that is, with density $e^{-x}, x>0$ and moment generating function $M_X(t)=\frac1{1-t}, t<1$. Then $\int_0^\infty M_X(-t)\; dt = \int_0^\infty \frac1{1+t} \; dt= \ln(1+t) \bigg\rvert_0^\infty = \infty$, so definitely do not converge, and is very different from $\frac1{\E X}=\frac11=1$.

Others have already explained that the answer to the question is NO, except trivial cases. Below we give an approach to finding $\DeclareMathOperator{\E}{\mathbb{E}} \E \frac1{X}$ when $X>0$ with probability one, and the moment generating function $M_X(t) = \E e^{tX}$ do exist. An application of this method (and a generalization) is given in Expected value of $1/x$ when $x$ follows a Beta distribution, we will here also give a simpler example.

First, note that $\int_0^\infty e^{-t x}\; dt = \frac1{x}$ (simple calculus exercise). Then, write $$ \E \left(\frac1{X}\right) = \int_0^\infty x^{-1} f(x)\; dx = \int_0^\infty \left( \int_0^\infty e^{-tx}\; dt \right) f(x)\; dx =\\ \int_0^\infty \left( \int_0^\infty e^{-tx} f(x) \; dx \right) \; dt = \int_0^\infty M_X(-t) \; dt $$ A simple application: Let $X$ have the exponential distribution with rate 1, that is, with density $e^{-x}, x>0$ and moment generating function $M_X(t)=\frac1{1-t}, t<1$. Then $\int_0^\infty M_X(-t)\; dt = \int_0^\infty \frac1{1+t} \; dt= \ln(1+t) \bigg\rvert_0^\infty = \infty$, so definitely do not converge, and is very different from $\frac1{\E X}=\frac11=1$.

1
source | link

Others have already explained that the answer to the question is NO, except trivial cases. Below we give an approach to finding $\DeclareMathOperator{\E}{\mathbb{E}} \E \frac1{X}$ when $X>0$ with probability one, and the moment generating function $M_X(t) = \E e^{tX}$ do exist. An application of this method (and a genaralization) is given in Expected value of $1/x$ when $x$ follows a Beta distribution, we will here also give a simpler example.

First, note that $\int_0^\infty e^{-t x}\; dt = \frac1{x}$ (simple calculua exercise). Then, write $$ \E \left(\frac1{X}\right) = \int_0^\infty x^{-1} f(x)\; dx = \int_0^\infty \left( \int_0^\infty e^{-tx}\; dt \right) f(x)\; dx =\\ \int_0^\infty \left( \int_0^\infty e^{-tx} f(x) \; dx \right) \; dt = \int_0^\infty M_X(-t) \; dt $$ A simple application: Let $X$ have the exponential distribution with rate 1, that is, with density $e^{-x}, x>0$ and moment generating function $M_X(t)=\frac1{1-t}, t<1$. Then $\int_0^\infty M_X(-t)\; dt = \int_0^\infty \frac1{1+t} \; dt= \ln(1+t) \bigg\rvert_0^\infty = \infty$, so definitely do not converge, and is very different from $\frac1{\E X}=\frac11=1$.