Say we have two random variables, x and y which are i.i.d. with common support [0, b]. I'm trying to evaluate the integral: $$ \int_{0}^{b}\int_{0}^{2A - x}f(y)\ dy \ f(x) \ dx $$

Where A is just some constant. I realise (well, I think) that this can be written as: $$ \int_{0}^{b}F(2A - x) \ f(x) \ dx $$

But I'm not quite sure how to interpret this (the expectation of P(x + y < 2A)?). Primary, I would wondering if it was possible to say something about it's relationship to F(A), i.e. greater than/less than under some condition on A and x. I thought it might be possible to say something using iterated expectations but I'm not sure.


Suppose that $X$ and $Y$ are independent random variables. Then, the the conditional pdf of $Y$ given the value of $X$ is the same as the unconditional pdf of $Y$: $$f_{Y\mid X}(y\mid X=x) = \frac{f_{X,Y}(x,y)}{f_X(x)} = \frac{f_X(x)f_Y(y)}{f_X(x)} = f_Y(y).$$ Now consider the computation of $P\{X+Y\leq 2A\}$ via the law of total probability expression \begin{align} P\{X+Y \leq 2A\} &= \int_{-\infty}^\infty P\{X+Y \leq 2A \mid X=x\} \cdot f_X(x)\, \mathrm dx\\ &= \int_{-\infty}^\infty P\{Y \leq 2A-x \mid X=x\} \cdot f_X(x)\, \mathrm dx\\ &= \int_{-\infty}^\infty \left[\int_{-\infty}^{2A-x} f_{Y \mid X=x}(y \mid X = x)\,\mathrm dy\right] \cdot f_X(x)\, \mathrm dx\\ &= \int_{-\infty}^\infty \left[\int_{-\infty}^{2A-x} f_{Y}(y)\, \mathrm dy\right] \cdot f_X(x)\, \mathrm dx \end{align}

For the special case when $X$ and $Y$ are also identically distributed and so $f_X(\cdot)=f_Y(\cdot) = f(\cdot)$, and also that $f(\cdot)$ has support $[0,b]$, the above simplifies to $$P\{X+Y \leq 2A\} = \int_{0}^b \left[\int_{0}^{2A-x}f(y)\, \mathrm dy\right] \cdot f(x)\, \mathrm dx. $$ There is, to the best of my knowledge, no simple expression relating the value of the above integral to $F(A)$. In particular, the claim in another answer that the value of the integral is $F(b)*F(2A-x)$ is incorrect. Note that $F(b) = 1$ and since the support is $[0,b]$ and what remains is a function of $x$ but $x$ has been "integrated out" in the double integral.

  • $\begingroup$ You are right, I should not treat the upper limit $x$ as a constant. I will delete the answer. $\endgroup$ – Deep North Jul 2 '17 at 3:47

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