# Multiplying or adding constants within $P(X \leq x)$?

Is it always true, that, for a positive constant $$c$$, we have $$P(X \leq x) = P(c X \leq cx)$$ for some continuous random variable $$X$$?

Further is it always true that $$P(X + c \leq x + c) = P(X \leq x)$$?

If so, what governs the equality?

When you write $$P(X \leq x)$$, hidden in the background (if you haven't learned measure-theoretic probability) is the following:

The actual meaning of $$P(X \leq x)$$ is: $$P(X \leq x) = P(\{\omega \in \Omega: X(\omega)\leq x\})\text{.}$$ This is regardless of whether $$X$$ is a continuous or discrete random variable.

The following facts are necessary to show that your equalities are true:

• Fact. If $$X$$ is a random variable, so is $$Y = X + c$$ for some constant $$c \in \mathbb{R}$$.
• Fact. If $$X$$ is a random variable, so is $$Z = cX$$ for some constant $$c \in \mathbb{R}$$.

Hence, we may see that \begin{align} P(Y \leq x + c) &= P(\{\omega: Y(\omega) \leq x + c\}) \\ &= P(\{\omega: (X + c)(\omega) \leq x + c\}) \\ &= P(\{\omega: X(\omega) +c\leq x + c\}) \end{align} which is clearly equal to $$P(X \leq x)$$.

One may show similarly for $$P(Z \leq cx)$$ for some $$c > 0$$.

You can do all those things.

$$X \le x$$ is an event in your probability space. To put it less formally: for each of your possible outcomes, you have an associated value of $$X$$, and $$X \le x$$ is the event (set of outcomes) corresponding to values of $$X$$ that are less than $$x$$.

If we change $$X \le x$$ to $$cX \le cx$$ (for $$c> 0$$) or $$X+c \le x+c$$ or anything that describes the same values of $$X$$, we have a different description of the same event. There are no outcomes which make $$X\le x$$ occur but don't make $$cX \le cx$$ occur, or vice versa. So its probability will be the same.

We can always do anything to the inside of $$\Pr[X \le x]$$ that we could do and undo to an inequality. We are looking for the two-way implication $$X \le x \iff X+c \le x+c$$. As long as that holds, we're good.

Often, we can actually do more manipulations with the inside of a probability, that are only valid because of our specific experiment. For example, if we flip lots of coins, and $$X$$ is the number that land heads, then $$X \le 2.5$$ and $$X \le 2$$ are different descriptions of the same event. Even though they are not equivalent statements, there are no outcomes for which one event occurs, and the other does not.

• "We can always do anything to the inside of $Pr[X \leq x]$ that we could do and undo to an inequality." This is my takeaway. I've been manipulating inequalities within a probability just like how I manipulate inequalities in algebra. I've always just assumed it's valid, but never really looked into what justifies it from a probabilistic standpoint. Commented May 8, 2021 at 14:01

Answer to both questions is yes. I'll come at this from a standpoint of not knowing anything about measure theory. For the first question, note that the area under the graph of a PDF must be $$1.$$ So we ask ourselves, what does the graph of the PDF of $$2X$$ look like compared to the graph of $$X$$? Well, it is spread out more by a factor of two in the $$x$$-direction. But if we left is at that, then the area under the PDF for $$2X$$ would be $$2$$. Therefore, we must make the following adjustment: if the PDF for the random variable $$X$$ is $$f(x)$$ with domain $$a\leq x \leq b,\$$ then the PDF for the random variable $$\ 2X\$$ is $$\frac{1}{2}f\left(\frac{1}{2}x\right)$$ with domain $$\ 2a\leq x \leq 2b.\$$ But $$2$$ was arbitrary here, and so we get the following more general result:

if the PDF for the random variable $$X$$ is $$f(x)$$ with domain $$-\infty\leq a\leq x \leq b\leq \infty,\$$ then for $$c>0,\$$ the PDF for the random variable $$\ cX\$$ is $$\frac{1}{c}f\left(\frac{1}{c}x\right)$$ with domain $$\ ca\leq x \leq cb.\$$

The diagram below demonstrates for $$c=2$$. 

The area of the two green rectangles is the same: the area of the rectangle in $$\frac{1}{2}f\left(\frac{1}{2}x\right),$$ is twice as short and twice as fat as it's corresponding rectangle on the $$f(x)$$ diagram. Thus "Area = $$1$$" has been preserved.

Since area is the limit of the sum of rectangles, these "approximate rectangles" can be formalised by the Lebesgue/measure theory. In fact, this is probably the purpose of Lebesgue/measure theory: to formalise this integration stuff.

Now, going back to your original question, which was:

is $$P(X \leq x_0) = P(c X \leq cx_0)$$ always true?

If we use some less technical results (than measure theory) like the fundamental theorem of calculus with the reverse chain rule, we can see that:

$$P(X \leq x_0) = \int_a^{x_0} f(x) dx = F(x_0) - F(a),$$

and

$$P(c X \leq cx_0) = \int_{ca}^{cx_0} \frac{1}{c}f\left(\frac{1}{c}x\right)dx = \left[\frac{1}{c}\times cF\left(\frac{1}{c}x\right)\right]_{ca}^{cx_0} = F(x_0) - F(a),$$

attaining the required result.

A similar way of thinking about shifting of graphs can be done to attain the affirmative result for the second question.

• Or maybe this line of reasoning is completely incorrect? If you add two Bimodal normal distributions, $X$, each with peaks at $-1$ and $1$, then $X+X$ will be a trimodal distribution with peaks at $-2, 0,$ and $2$. So this answer is wrong then? Commented May 8, 2021 at 15:22
• What's a "bimodal normal distribution"? (Also, the distribution of $2X$ is not the same as the distribution of the sum of two independent copies of $X$.) Commented May 9, 2021 at 22:38
• Like a Normal Distribution but with two peaks instead of one. See: en.m.wikipedia.org/wiki/Multimodal_distribution Also, are you sure that is true? I thought $2X$ and $X+X$ were the same (although I’m not certain)? Commented May 9, 2021 at 22:39
• $2X$ and $X+X$ are the same, but $X+X$ is not the same as adding together two independent copies of $X$. Consider a simple distribution which is $0$ or $1$ with equal probability. If $X$ has this distribution, then both $X+X$ and $2X$ will be $0$ or $2$ with equal probability. But if $X_1, X_2$ have this distribution, then $X_1 + X_2$ will be $0, 1, 2$ with probabilities $\frac14, \frac12, \frac14$ respectively. Commented May 9, 2021 at 22:53