# On an exercise in Finding the cdf of a random variable given the pdf of an other random variable.

This exercise is taken from Casella and Berger Statistical Inference, it's number 1.55.

I state the text: An electronic device has lifetime denoted by $T$. The device has value $V=5$ if it fails before time $t = 3$; otherwise, it has value $V = 2T$. Find the cdf of $V$, if $T$ has pdf

$$f_T ( t) = \frac{2}{3} e^\frac{-2t}{3}, \quad t >0.$$

Could someone walk me through the steps? I could do the other exercises but I am a bit confused on the dependence of one random variable on the other.

• The use of the word "value" is confusing to me. Does this indicate that $V$ "has the same distribution as" ...? Feb 25 '16 at 20:43
• To me, this question doesn't seem to make much sense that way... Feb 25 '16 at 22:32

$V$ is a function of the random variable $T$, and the behavior of $V$ changes when $T=3$. So to calculate the probability $P(V\le v)$ you should argue differently depending on whether $T\le3$ or $T>3$. This means you break up the probability into two pieces: $$P(V\le v)=P(V\le v, T\le3) + P(V\le v, T>3)\tag1$$ To decide how to proceed from here, look at what values $V$ can take: It has a mass at value $5$, and values from $6$ onward. So to cover all possibilities, you should consider four cases: (a) $v<5$ (b) $v=5$ (c) $5< v<6$ (d) $v \ge 6$.
For case (a), the second term on the RHS of (1) is zero because $V>6$ when $T>3$ so it's impossible for $V\le v$. The first term is also zero because $V=5$ when $T\le3$.
For case (b) and (c), the second term is zero using the same argument as in (a). You should get the first term to be $P(T\le3)$, because $V=5$ when $T\le3$, so for sure $V\le v$.
For case (d), the first term on the RHS of (1) simplifies to $P(T\le3)$ using the same reasoning as in the previous case. The second term simplifies to $P(2T\le v,T>3)$ because $V=2T$ when $T>3$. Therefore (1) becomes $$P(V\le v)=P(T\le3)+P(3<T\le v/2)=P(T\le v/2).$$