# Thread:

1. ## Distribution function for a function of random variable

Suppose that X is a random variable with distribution function $F_X(t)$ and let $Y=a+bX \ \ \ b<0$. Derive the distribution function for Y.

My solution:

I know that $F_Y(t)=P(T \leq t)=P(a+bX \leq t)$

$=P(X \geq \frac {t-a} {b} ) =1 - P( X \leq \frac {t-a}{b})=1- F_X ( \frac {t-a}{b} )$

But the answer is wrong, why?

Thanks.

2. Hello,

More precisely, it's $1-\mathbb{P}\left(X{\color{red}<} \frac{t-a}{b}\right)$

There are some situations (in particular for some discrete random variables) where $\mathbb{P}\left(X< \frac{t-a}{b}\right)\neq \mathbb{P}\left(X\leq \frac{t-a}{b}\right)$

3. For some reason, the back of the book says the answer is:

$1-F_X( \frac {t-a}{b})+p_X ( \frac {t-a}{b})$

4. Originally Posted by tttcomrader
For some reason, the back of the book says the answer is:

$1-F_X( \frac {t-a}{b})+p_X ( \frac {t-a}{b})$
So that confirms what Moo suggested...

You have $1 - P( X < \frac {t-a}{b})$ So you just need to add the case $P( X = \frac {t-a}{b})$
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