grandes-ecoles 2018 Q6

grandes-ecoles · France · centrale-maths1__mp Continuous Probability Distributions and Random Variables Expectation and Moment Inequality Proof

Let $p$ and $q$ be two strictly positive reals such that $\frac{1}{p} + \frac{1}{q} = 1$. Deduce that if $X$ and $Y$ are two real-valued random variables on the finite probability space $(\Omega, \mathcal{A}, \mathbb{P})$, then
$$\mathbb{E}(|XY|) \leqslant \mathbb{E}(|X|^{p})^{1/p} \mathbb{E}(|Y|^{q})^{1/q}$$
You may first prove this result when $\mathbb{E}(|X|^{p}) = \mathbb{E}(|Y|^{q}) = 1$.

Let $p$ and $q$ be two strictly positive reals such that $\frac{1}{p} + \frac{1}{q} = 1$. Deduce that if $X$ and $Y$ are two real-valued random variables on the finite probability space $(\Omega, \mathcal{A}, \mathbb{P})$, then

$$\mathbb{E}(|XY|) \leqslant \mathbb{E}(|X|^{p})^{1/p} \mathbb{E}(|Y|^{q})^{1/q}$$

You may first prove this result when $\mathbb{E}(|X|^{p}) = \mathbb{E}(|Y|^{q}) = 1$.

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