# Questions tagged [probability-inequalities]

Probability Inequalities are useful for bounding quantities that might otherwise be hard to compute. A related concept is a concentration inequality, which specifically provides bounds on how far a random variable deviates from some value.

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### Concentration for Conditional Random Variable

Consider a conditional random variable \begin{equation} X = \begin{cases} Y & \quad\quad ,X \in A \\ Z & \quad\quad ,X \in A^\complement \end{cases} \end{equation} $Y$ ...
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### Zero-mean RV $X$, probability of being positive using moments

For zero-mean RV $X$ with finite fourth moment, prove that $$P(X>0)\ge \frac{\mathbb{E}(X^2)^2}{4\mathbb{E}(X^4)}$$ I tried Chebyshev with adding $t$ to both sides, but I could not get fourth ...
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### Convergence of a sequence in finite number of steps

Here is the setup of my problem. It is a sequential problem and there are two possible actions A and B. Now, when either action $A$ or $B$ is taken at the $j$th time point, we observe some outcome say ...
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### Characteristic function inequality

Random variable $X$ and its characteristic function $\phi_X(t)$ then $$\Pr\left(|X|>\frac2T\right) \leq 2\left(1 - \frac1{2T}\int_{-T}^{T}\phi_X(t)dt\right)$$ I cannot find a way how to ...
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### Why does conditional expectation have this property for independent random variables?

For a reference, please see pp. 53-54 of Boucheron, Lugosi, Massart, Concentration Inequalities: A Nonasymptotic Theory of Independence. Let $f: \mathcal{X}^n \to \mathbb{R}$ be a measurable function (...
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### Increased probability of event during period of time

In game of FIFA there are packs by opening which a user receives soccer player cards. The higher the rating of a player card the rarer it drops depending on some kind of random number generator. Since ...
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### Linear combination of truncated normals

I am trying to calculate the following expression: $$Z = \mathbb{E}\left[\left| \langle \mathbf{a}, u \rangle \right| \right] = \left| \sum_{i=1}^d a_i u_i \right|, \quad \left\| u \right\| = 1$$ ...
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### Proof that $E(|X_1 - X_2|)$ is bound by twice the mean

Let $X_1$, $X_2$ be iid random variables. How do I show that for non-negative variables $E(|X_1 - X_2|)$ is bound from above by twice the expected value of $X_1$ (or $X_2$)?
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