# Obtaining different CDF of non-standard normal distribution using two methods

I was reading about the derivation of the CDF of a non-standard normal distribution.

One particular version goes like this:

Let $$X \sim \mathcal{N}(\mu, \sigma^2), Z \sim \mathcal{N}(0,1).$$

$$\therefore X = \sigma Z + \mu$$

$$P(X\le x) = P\left(\frac{X-\mu}{\sigma} \le \frac{x-\mu}{\sigma}\right) = \Phi\left(\frac{x-\mu}{\sigma}\right)$$

Now, I thought of another variation, this time arriving at a different conclusion:

$$P(X\le x) = P\left(\frac{X-\mu}{\sigma} \le \frac{x-\mu}{\sigma}\right) = P(Z\le z)$$

My version looks grossly incorrect. How could the CDF of $$X$$ and $$Z$$ be the same? Could someone please advise me on what I am doing wrong?

They're not the same because $$z$$ isn't $$x$$, but rather is $$(x-\mu)/\sigma$$.

• Yeah. I understand that the CDFs ought to be different. But I cannot understand why I managed to derive an equality between them in my working (which I think is mathematically correct). Did I intepret my own conclusion wrongly? Commented Jul 28, 2019 at 6:48
• @LanceHAOH You've shown a function of $x$ is equal to a function of $z$, with $z$ being some function of $x$ other than the identity function. But the functions aren't "the same" in the sense of returning the same output for a given shared input. It's like how $f(y):=y^9,\,g(y):=y^3\implies f(y)=g(y^3)$.
– J.G.
Commented Jul 28, 2019 at 7:01
• Ok. I understand your example($f \neq g$ although their values are equal given different inputs). The thing that is confusing me is this thought: "CDF of $X$ = $P(X \le x)$ = $P(Z \le z)$ = CDF of $Z$". I am not sure whether my example is valid though. Would you mind helping to clarify this? Commented Jul 28, 2019 at 7:26
• @LanceHAOH Since $X\le x$ is equivalent to $Z\le z$ for $z=\frac{x-\mu}{\sigma}$, the two conditions indeed have equal probabilities. But $X,\,Z$ have different CDFs because e.g. $P(X\le 1)\ne P(Z\le 1)$.
– J.G.
Commented Jul 28, 2019 at 7:30
• @LanceHAOH That's exactly right. What's more, this isn't a unique characteristic of linear transformations among Normal distributions; you could pull the same trick with any order-preserving transformation of any probability distribution.
– J.G.
Commented Jul 28, 2019 at 9:56

In their working $$\Phi$$ is CDF of $$Z$$.

In your working, you did not define what is $$z$$. I believe you want to define $$z$$ such that $$z=\frac{x-\mu}{\sigma}.$$

We have $$P(X \le x) =P(Z \le z)= P(Z \le \frac{x-\mu}{\sigma}).$$

• "you did not define what is $z$" Hmm. $z$ is just the output of the function $Z$ right? Since $Z=\frac{X-\mu}{\sigma}$, shouldn't $z=\frac{x-\mu}{\sigma}$ where $x$ is the output of the function $X$? Commented Jul 28, 2019 at 7:12
• $z$ is just a symbol of which I can guess the intended meaning. You might let to include a linear let $z=\frac{x-\mu}{\sigma}$ explicitly. Simply because you have defined $Z$, it doesn't mean that $z$ has been defined. Commented Jul 28, 2019 at 7:22