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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 !

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  • $\begingroup$ Could you write your question in your post ? $\endgroup$
    – Canardini
    Dec 17 '16 at 4:30
  • $\begingroup$ I have no ? Here it is anyway :) $\endgroup$ Dec 17 '16 at 13:33
  • $\begingroup$ 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? $\endgroup$ Dec 17 '16 at 13:36
  • $\begingroup$ @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 $. $\endgroup$
    – Michael R
    Dec 18 '16 at 4:30
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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 $.

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

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