Does the conditional density $f(X|X+Y<a)$ equal to $f(X)$ when $X,Y$ are independent continuous random variables with $Y$ having full support?

Question

I wonder if the following equation holds when continuous random variables $X,Y$ are independent from each other and random variable $Y$ has full support on $\mathbb{R}$. \begin{align} f(X)=f(X|X+Y<a) \end{align} where $a$ is a real constant.

Answer 1

Imagine a plot of the joint distribution of $X,Y$. The condition $X+Y <a$ can be depicted by a diagonal line $y=a-x$ as a boundary. The points below it satisfy the condition.

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Second image

Next, consider the marginal distribution by summing all the distributions of $X$ for the various values of $Y$. This will be a sum of right truncated distributions $ P_{X\vert X<c}(x\vert c) $ of $X$ (Three of those truncated distributions are shown in the second image in red, green and blue. Also the related points of the joint distribution are coloured in the first graph).

Your end result will be

$$\sum_{\forall y} P_{X\vert X <c}(x\vert a-y) P_Y (y) $$

When $P_Y (y)$ is positive everywhere then it must be that you are taking some bite out of the right side of the distribution of $X $.

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Third image

You could also view the image in another direction. And quantify how much of a bite you take out of $X $ by using the cdf $P_{Y<c}(c)$

$$P_{X|X+Y <a}(x|y,a) \propto P_X(x) \cdot P_{Y <c}(a-x) $$

When The PDF of $Y $ is positive everywhere (full support), then the CDF $P_{Y <c}(y)$ will be a continuously increasing function and the 'bite' will be larger for higher values of $X$. (you can see this in the image as the increasing fraction of gray points - those that do not fulfill the condition - as you move to higher for values of $x$)

$X $ remaines only unaffected when $X $ follows a degenerate distribution (ie has only one value). (or when $Y$ has not full support, and more particularly when the CDF of $Y$ is constant for all values of $X$, ie when $P_Y(z) = 0$ if $P_X(z)>0$ )

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The above formulas are expressed for discrete variables but the same logic works for continuous variables.

$$f_{X \vert X+Y<a}(x \vert a) \propto f_X(x) F_Y(a-x)$$