# Inequality with products of expectations of functions of continuous random variables

I am trying to solve the following problem:

Suppose $X$ is a continuous random variable with density $f$ over $[0,1]$. Then show that $(\int_0^1xf(x)dx)(\int_0^1x^3f(x)dx)>(\int_0^1x^2f(x)dx)^2$.

I am trying to apply Fubini's theorem, but am getting equality for the two sides. What mistake am I making?

• Now sure about the strict inequality, but for the weak inequality you may utilize the Cauchy-Schwarz inequality. – Sangchul Lee Feb 19 '18 at 6:13
• The inequality is strict as in CS equality is only when the integrands in LHS are proportional. – Macavity Feb 19 '18 at 8:34

Rewriting the integrals and applying Holder's inequality.

Notice:

$(\int_0^1xf(x)dx)(\int_0^1x^3f(x)dx)= (\int_0^1xf(x)dx)^{\frac{1}{2}}\cdot(\int_0^1x^3f(x)dx)^{\frac{1}{2}}\cdot(\int_0^1xf(x)dx)^{\frac{1}{2}}\cdot(\int_0^1x^3f(x)dx)^{\frac{1}{2}}$

look at $(\int_0^1xf(x)dx)^{\frac{1}{2}}(\int_0^1x^3f(x)dx)^{\frac{1}{2}}$:

$$(\int_0^1xf(x)dx)^{\frac{1}{2}}(\int_0^1x^3f(x)dx)^{\frac{1}{2}} = (\int_0^1(x^{\frac{1}{2}})^2f(x)dx)^{\frac{1}{2}}(\int_0^1(x^{\frac{3}{2}})^2f(x)dx)^{\frac{1}{2}}$$

By Holder (with $d\mu = f(x)dx$):

$$(\int_0^1(x^{\frac{1}{2}})^2f(x)dx)^{\frac{1}{2}}(\int_0^1(x^{\frac{3}{2}})^2f(x)dx)^{\frac{1}{2}} \geq (\int_0^1x^{\frac{1}{2}}x^{\frac{3}{2}}f(x)dx) = (\int_0^1x^2f(x)dx)$$

Because

$$(\int_0^1xf(x)dx)(\int_0^1x^3f(x)dx) = \{(\int_0^1xf(x)dx)^{\frac{1}{2}}(\int_0^1x^3f(x)dx)^{\frac{1}{2}}\}^2$$

$$(\int_0^1xf(x)dx)(\int_0^1x^3f(x)dx) \geq (\int_0^1x^2f(x)dx)^2$$

• Thanks all. I am still struggling to prove the strict inequality, which I think probably holds. Certainly I have not found a counter example. – banpriyo Feb 19 '18 at 8:37

Since $f(x)\geq0$, I think it's just C-S: $$\int\limits_0^1xf(x)dx\int\limits_0^1x^3f(x)dx\geq\left(\int\limits_0^1\sqrt{xf(x)x^3f(x)}dx\right)^2=\left(\int\limits_0^1x^2f(x)dx\right)^2.$$