Expected value of a function of a random variable

[AllRelative]

Homework Statement

Let X be a random variable. It is not specified if it is continuous or discrete. Let g(x) alway positive and strictly increasing. Deduce this inequality:
$$P(X\geqslant x) \leqslant \frac{Eg(X)}{g(x)} \: $$
where x is real.

Homework Equations

I know that if X is discrete
$$E(X) =\sum_{n=1}^{\infty} g(x_i)x_i$$

And if X is continuous,
$$\int_{-\infty}^{\infty} g(x)f(x) dx$$

The Attempt at a Solution

Is there a way to answer the question without proving the two cases (cont and discrete). Thanks!

 

[StoneTemplePython]

The Attempt at a Solution

Is there a way to answer the question without proving the two cases (cont and discrete). Thanks!

Yes -- use indicator random variables and recognize that your problem is actually asking for a standard proof of Markov's Inequality which takes just one or two lines.

Note: if you want something more concrete, let $g$ be the exponential function.

 

[AllRelative]

Yes -- use indicator random variables and recognize that your problem is actually asking for a standard proof of Markov's Inequality which takes just one or two lines.

Note: if you want something more concrete, let $g$ be the exponential function.

I just read up of Markov's inequality. I see that it is the same problem except for the function. I'm just unsure about what to do with the function g.

$$g(x) P(X)\geq E(g(I_{X\geq x}))$$

How does an idicator function behave in a function? That's what confuses me. If g wasn't there I could finish the proof...

 

[StoneTemplePython]

I just read up of Markov's inequality. I see that it is the same problem except for the function. I'm just unsure about what to do with the function g.

$$g(x) P(X)\geq E(g(I_{X\geq x}))$$

How does an idicator function behave in a function? That's what confuses me. If g wasn't there I could finish the proof...

some of the things look backwards here?
- - - -
Suggestion: break it into two parts.

(part 1) let random variable $Y := g(X)$. Now prove markov's inequality for $Y$.

(part 2) after you have done the above, reason through -- how can you relate Y and X? i.e. if I say an experiment occurs in the sample space and I know the outcome is $Y(\omega) \geq g(c)$ that immediately tells you something of interest about $X(\omega)$. Again, de-abstracting this and having $g$ be the exponential function may be useful for now...

 

[AllRelative]

Oh right... Thanks man!

 

[Ray Vickson]

Homework Statement

Let X be a random variable. It is not specified if it is continuous or discrete. Let g(x) alway positive and strictly increasing. Deduce this inequality:
$$P(X\geqslant x) \leqslant \frac{Eg(X)}{g(x)} \: $$
where x is real.

Homework Equations

I know that if X is discrete
$$E(X) =\sum_{n=1}^{\infty} g(x_i)x_i$$

And if X is continuous,
$$\int_{-\infty}^{\infty} g(x)f(x) dx$$

The Attempt at a Solution

Is there a way to answer the question without proving the two cases (cont and discrete). Thanks!

Now that you have done the question I can show you may favorite, quick way of doing it. For $a > 0$ let $v_a(x) = g(x)/g(a)$ and let
$$u_a(x) = 1\{ x \geq a \} = \begin{cases} 0 & \text{if} \; x < a \\
1 & \text{if} \; x \geq a
\end{cases}$$
For all $x \geq 0$ we have $0 \leq u_a(x) \leq v_a(x)$, with $u_a(a) = v_a(a) = 1.$ Thus
$$P(X \geq a) = E u_a(X) \leq E v_a(X) = E g(X)/g(a).$$

This just uses elementary properties of expectation, and works the same way whether $X$ is discrete, continuous, or mixed discrete-continuous.

Here is a drawing that shows the situation.