# Discrete Random Variable Transformation

i have a problem and i can't figure out any solution.

Suppose i have this game: i throw a die untill i get a 6. Every time i throw the dice i pay -1 and when i get the 6 i win 5. (Nb: when i obtain the first 6 the game end).

I call $X$ a radom variable that is the number of attempts to get a 6. $X$ has a geometric distribution with parameter $\theta = \frac{1}{6}$.

Now define a variable that tells me how much i win/loss: $Y = 5 - X$; this is my transformation.

The problem is that i don't know how to compute the Cumulative distribution function for the $Y$.

I tried: $F_Y(y) = P(Y \le y) = P(5 - X \le y) = P(X \ge y - 5) = 1 - P(X < y-5) = 1 - F_X(y -5 -1)$ sustituting: $F_Y(y) = 1 - (1 - (\frac{5}{6})^{y - 6}) = (\frac{5}{6})^{y - 6}$

That is wrong! Can you help me finding the mistake?

Thank you, bye.

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The singular of "dice" is "die". –  joriki Mar 21 '12 at 11:41
This looks a lot like homework, and if so, please add the homework tag. It seems that it would be easier to find (i) the possible values of $Y$, (ii) the probabilities that $Y$ takes on each of these values, and (iii) adding up $P\{Y = n\}$ for all $n \leq y$ to get $F_Y(y)$. Note that $F_Y(y)$ is defined for all real numbers $y$, not just for integer values of $y$. –  Dilip Sarwate Mar 21 '12 at 11:45
Yes, maybe in this specific case it is easier...but i'm searching for a general procedure to solve this kind of problems.. –  Aslan986 Mar 21 '12 at 11:53
$$\mathbb{P}(Y \leqslant y) = \mathbb{P}(5-X \leqslant y) \stackrel{\color\red{\text{!!!}}}{=} \mathbb{P}(X \geqslant 5-y) = 1-\mathbb{P}(X < 5-y) = 1 - F_X(4-y)$$ Compare this with your answer $1-F_X(y-6)$.
OkOk Sasha i got it, thank you! Just a correction..i think it is: $1 - F_X(4 - y)$ –  Aslan986 Mar 21 '12 at 14:27