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?

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 where $$a$$ is a real constant.

• Consider $X, Y$ i.i.d. Normal$(0,1)$, and $a=-3$. Is the probability that $X>0$ still $50\%$? – jbowman Sep 19 '19 at 1:49
• @jbowman Thank you. I feel the equation does not hold. But, on the other hand, I always feel that I can vary $Y$ freely to accommodate any $X$. So, the inequality is not really a restriction. I know this should be wrong, but I don't know where I am wrong intuitively. – zxjroger Sep 19 '19 at 2:20

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

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 $$P_{X\vert X of $$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).

$$\sum_{\forall y} P_{X\vert X

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 $$X$$.

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

$$P_{X|X+Y

When The PDF of $$Y$$ is positive everywhere (full support), then the CDF $$P_{Y will be a continuously increasing function and the 'bite' will be larger for higher values of $$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 $$P_X(z)>0$$ )

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

$$f_{X \vert X+Y

• Thank you so much! I read it very carefully. This is super clear and helpful. – zxjroger Sep 19 '19 at 15:23