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Note: I will use $\mathbb{P}\{X \in dx\}$ to denote $f(x)dx$ where $f(x)$ is the pdf of $X$.

While doing some homework, I came across a fault in my intuition. I was scaling a standard normally distributed random variable $Z$. Edit: I was missing the infinitesimals $dx$ and $dx/c$, so everything works out in the end. Thank you Jokiri! $$\mathbb{P}\{cX \in dx\} = \frac{e^{-x^2 / 2c^2}}{\sqrt{2\pi c^2}}$$ while $$\mathbb{P} \left\{X \in \frac{dx}{c}\right\} = \frac{e^{-(x/c)^2/2}}{\sqrt{2\pi}}$$

Could anyone help me understand why the following equality doesn't hold? Edit: it does, see edit below $$\mathbb{P}\{cX \in dx\} \ne \mathbb{P} \left\{X \in \frac{dx}{c}\right\}$$

I have been looking around, and it seems that equality of the cdf holds, though: $$\mathbb{P}\{cX < x \} = \mathbb{P}\left\{X < \frac xc\right\}.$$

Thank you in advance!


This question came out of a silly mistake on my part. Let me attempt to try to set things straight.

Let $Y = cX$. Let $X$ have pdf $f_X$ and $Y$ have pdf $f_Y$.

$$\mathbb{E}[Y] = \mathbb{E}[cX] = \int_{-\infty}^\infty cx\,f_X(x)\mathop{dx} =\int_{-\infty}^\infty y\, f_X(y/c) \frac{dy}{c}$$

So, $f_Y(y) = \frac 1c f_X(y/c)$.

Thank you for the help, and sorry for my mistake.

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Typo last line, presumably you want $X\lt \frac{x}{c}$. Unfortunately cannot write an answer to your question, because of unwillingness to use the notation. –  André Nicolas Jan 6 '13 at 1:26
    
Thanks for catching the typo. What notation do you prefer? Maybe $\mathbb{P}\{X=x\}$ instead? –  angryavian Jan 6 '13 at 1:55
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As much as possible, prefer to work with cdf, which has a concrete probabilistic meaning. –  André Nicolas Jan 6 '13 at 1:56
    
I strongly agree with André. Please do not "introduce" notation that does not make sense. –  TMM Jan 6 '13 at 2:40
    
I apologize, this was the notation that I was taught to use. I would really like to know what the preferred notation is. Using $f(x)$ as the pdf? –  angryavian Jan 6 '13 at 2:47
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2 Answers 2

up vote 1 down vote accepted

Setting aside rigour and following your intuition about infinitesimal probabilities of finding a random variable in an infinitesimal interval, I note that the left-hand sides of your first two equations are infinitesimal whereas the right-hand sides are finite. So these are clearly wrong, even loosely interpreted. They make sense if you multiply the right-hand sides by the infinitesimal interval lengths, $\mathrm dx$ and $\mathrm dx/c$, respectively, and then everything comes out right.

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Thank you, sorry about the confusion. I hope everything is clear now! –  angryavian Jan 6 '13 at 2:09
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As you said yourself, $\mathbb P \left\{ X \in dx \right \} = f(x)dx$ so $$\mathbb{P} \left\{cX \in dx\right\} = \frac{e^{-x^2/2c^2}}{\sqrt{2\pi c^2}}dx$$

and

$$\mathbb{P} \left\{X \in \frac{dx}{c}\right\} = \frac{e^{-(x/c)^2/2}}{\sqrt{2\pi}}\frac{dx}{c}=\mathbb{P} \left\{cX \in dx\right\}$$

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