Probability for random variable that depends on other random variables (deferred decision)

Question

I read a proof which used the following type of argument:

Assume we draw two random variables $x_1, x_2$ indepenently. Let $y$ be another random variable, that directly depends on $x_1$ and $x_2$ (e.g. $y$ is the sum of both). Suppose that under the assumption, that we have drawn $x_2$ already, the bound $\mathbb{P}(y \leq \varepsilon) \leq \frac{\alpha}{x_2}$ is true. Let $f$ be the pdf for $x_2$. We then get $$ \mathbb{P}(y \leq \varepsilon)\leq \int_0 ^\infty \frac{\alpha}{x_2} f(x_2) \ dx_2 .$$

Intuitively this makes so much sense, but I dont know why exactly this is true. What is the underlying probability theoretic principle that allows us to conclude the second bound?

Answer 1

Let me first clean up the notation in your proof. I am used to writing random variables as capital bold letters so they're easily distinguishable from non-random quantities. I don't insist on that convention, but it is important to distinguish a random variable from a possible value it can have. In particular, it makes no sense to integrate with respect to a random variable!

Assume we draw two random variables $\mathbf X_1, \mathbf X_2$ indepenently. Let $\mathbf Y$ be another random variable, that directly depends on $\mathbf X_1$ and $\mathbf X_2$ (e.g. $\mathbf Y$ is the sum of both). Suppose that under the assumption that we have drawn $\mathbf X_2$ already, the bound $\mathbb{P}(\mathbf Y \leq \varepsilon \mid \mathbf X_2 = x_2) \leq \frac{\alpha}{x_2}$ is true. Let $f$ be the PDF for $\mathbf X_2$. We then get $$ \mathbb{P}(\mathbf Y \leq \varepsilon)\leq \int_0 ^\infty \frac{\alpha}{x_2} f(x_2) \ dx_2.$$

Here, we can use the law of total probability for continuous random variables: $$ \mathbb P(A) = \int_{-\infty}^\infty \mathbb P(A \mid \mathbf X = x) f_{\mathbf X}(x) \,dx. $$ In this problem $A$ will be the event $\mathbf Y \le \varepsilon$ and $\mathbf X$ will be $\mathbf X_2$. We should assume that $\mathbf X_2$ can never take on negative values, or else the bound in the question can't possibly hold; therefore it is okay to integrate from $0$ to $\infty$ instead. This gives us $$ \mathbb P(\mathbf Y \le \varepsilon) = \int_0^\infty \mathbb{P}(\mathbf Y \leq \varepsilon \mid \mathbf X_2 = x_2) f(x_2) \,dx_2. $$ Using the inequality $\mathbb{P}(\mathbf Y \leq \varepsilon \mid \mathbf X_2 = x_2) \leq \frac{\alpha}{x_2}$, we get the final bound you wanted.