# Suppose that Y is a r.v. with CDF t^3+1 / 2 -1<=t<= 1. Find the dist function for w=8+2y and it's pdf and what is the median of w?

So, I know that since I have $F_y(t)$ I can say that: $$F_z(t) = P(Z\leq t) = P (8+2Y\leq t) = P (Y <= t-8/2) = F_y (t-8/2)$$

From this I have the distribution function so am I just differentiating $(t-8/2)^3$ then to get the pdf ?

Also I know the median is $P(Y\leq t)=0.5$ so previously I done that $t-8/2 = 0.5$ but again I am totally unsure and pretty sure it's wrong.

Any help would be great !

• Could you write your question in your post ? Dec 17 '16 at 4:30
• I have no ? Here it is anyway :) Dec 17 '16 at 13:33
• I have no? Here it is again :) Suppose that Y is a random variable with cumulative distribution function FY (t) = 0, t < −1 t^3+1/2 , −1 ≤ t ≤ 1 1, t > 1 (a) Find the distribution function for W = 8 + 2Y and the density function for W. (b) What is the median of W? Dec 17 '16 at 13:36
• @XeroPhobous I answered your question below. Please let me know if you have any questions. However, I think that there is a mistake in the CDF that you provided... Instead of $F_Y(t) = t^3 + \frac{1}{2}$, it should be $F_Y(t) = \frac{t^3 + 1}{2}$. The CDF should be equal to 1 at $t = 1$ and 0 at $t = -1$. Dec 18 '16 at 4:30

You are on the right track. You want to write the CDF as follows:

$F_W(t) = F(W \leq t) = P(8+2Y \leq t) = P \Big(Y \leq \frac{t-8}{2} \Big) = F_Y \Big( \frac{t-8}{2} \Big)$

Next, you want to compute $f_W(t) = \frac{dF_W(t)}{dt} = \frac{d}{dt} F_Y \Big( \frac{t-8}{2} \Big)$.

Case 1 $(t<-1):$

Since $F_Y(t) = 0$ when $t < -1$, then $\frac{d}{dt} F_Y \Big( \frac{t-8}{2} \Big) = 0$ when $\frac{t-8}{2}<-1 \rightarrow t < 6$. So $f_W(t) = 0$ when $t < 6$.

Case 2 $(t>1):$

Since $F_Y(t) = 1$ when $t > 1$, then $\frac{d}{dt} F_Y \Big( \frac{t-8}{2} \Big) = 0$ when $\frac{t-8}{2}>1 \rightarrow t > 10$. So $f_W(t) = 0$ when $t < 10$.

Case 3 $(-1 \leq t \leq 1):$

Since $F_Y(t) = \frac{t^3 + 1}{2}$ when $(-1 \leq t \leq 1)$, then $F_Y \Big(\frac{t-8}{2} \Big) = \frac{1}{2} \Big( \Big( \frac{t-8}{2} \Big)^3 + 1 \Big)$ when $-1 \leq \frac{t-8}{2} \leq 1 \rightarrow 6 \leq t \leq 10$. Finally, we want to differentiate with respect to t, i.e.

$f_W(t) = \frac{dF_W(t)}{dt} = \frac{d}{dt} F_Y \Big( \frac{t-8}{2} \Big) = \frac{d}{dt} \Big( \frac{1}{2} \Big( \frac{t-8}{2} \Big)^3 + \frac{1}{2} \Big) = \frac{3}{2} \Big( \frac{t-8}{2} \Big)^2 \frac{1}{2} = \frac{3}{4} \Big( \frac{t-8}{2} \Big)^2$

Median:

Since the median is the point $t_0$ where $F_W(t_0) = \frac{1}{2}$ you can solve for it as follows:

$F_W(t) = \frac{1}{2} \rightarrow F_W(t) = F_Y \Big( \frac{t-8}{2} \Big) = \frac{1}{2} \Big( \frac{t-8}{2} \Big)^3 + \frac{1}{2} = \frac{1}{2} \rightarrow t = 8$.

• Yep sorry about the edit you were correct ! Thanks a lot, very easy to get once it's explained properly ! Dec 20 '16 at 4:39
• @XeroPhobous You are welcome! If you don't have any more questions, could you please accept the answer? Thanks! Dec 20 '16 at 13:27