Question

Let \( X_{i} \sim \operatorname{Exp}(1), i=1,2, \ldots \), independently. Show that \( Y=X_{1}+X_{2} \) is not an exponential r.v. Hint: Recall from class when we solved \#8.2. Let \( f_{1}\left(x_{1}\right) \) and \( f_{2}\left(x_{2}\right) \) denote the exponential p.d.f. for \( X_{1} \) and \( X_{2} \), respectively. Which of the following statements is a valid argument? Mark the correct choice. (a) \( f_{1}\left(x_{1}\right)=e^{-x_{1}} \) and \( f_{2}\left(x_{2}\right)=e^{-x_{2}} \Longrightarrow f_{Y}(y)=e^{-x_{1}}+e^{-x_{2}} \) (b) \( f_{Y}(y)=e^{f_{1}\left(x_{1}\right)+f_{2}\left(x_{2}\right)}=e^{f_{1}\left(y-x_{2}\right)+f_{2}\left(y-x_{1}\right)} \) (c) \( f_{Y}(y)=\int_{0}^{y} e^{-\left(y-x_{1}\right)} e^{-x_{1}} d x_{1}=y e^{-y} \) (d) Use the change of variable formula, \( f_{Y}(y)=f_{1}\left(y-x_{2}\right)\left|\frac{d x}{d_{y}}\right|=\epsilon^{-\left(y-x_{2}\right)} \) (e) none of these

          Let \( X_{i} \sim \operatorname{Exp}(1), i=1,2, \ldots \), independently.
Show that \( Y=X_{1}+X_{2} \) is not an exponential r.v.
Hint: Recall from class when we solved \#8.2. Let \( f_{1}\left(x_{1}\right) \) and \( f_{2}\left(x_{2}\right) \) denote the exponential p.d.f. for \( X_{1} \) and \( X_{2} \), respectively.
Which of the following statements is a valid argument? Mark the correct choice.
(a) \( f_{1}\left(x_{1}\right)=e^{-x_{1}} \) and \( f_{2}\left(x_{2}\right)=e^{-x_{2}} \Longrightarrow f_{Y}(y)=e^{-x_{1}}+e^{-x_{2}} \)
(b) \( f_{Y}(y)=e^{f_{1}\left(x_{1}\right)+f_{2}\left(x_{2}\right)}=e^{f_{1}\left(y-x_{2}\right)+f_{2}\left(y-x_{1}\right)} \)
(c) \( f_{Y}(y)=\int_{0}^{y} e^{-\left(y-x_{1}\right)} e^{-x_{1}} d x_{1}=y e^{-y} \)
(d) Use the change of variable formula, \( f_{Y}(y)=f_{1}\left(y-x_{2}\right)\left|\frac{d x}{d_{y}}\right|=\epsilon^{-\left(y-x_{2}\right)} \)
(e) none of these

Let Xi∼Exp(1), i=1,2, …, independently.
Show that Y=X1+X2 is not an exponential r.v.
Hint: Recall from class when we solved #8.2. Let f1(x1) and f2(x2) denote the exponential p.d.f. for X1 and X2, respectively.
Which of the following statements is a valid argument? Mark the correct choice.
(a) f1(x1)=e^-x1 and f2(x2)=e^-x2⟹ fY(y)=e^-x1+e^-x2
(b) fY(y)=e^f1(x1)+f2(x2)=e^f1(y-x2)+f2(y-x1)
(c) fY(y)=∫0^y e^-(y-x1) e^-x1 d x1=y e^-y
(d) Use the change of variable formula, fY(y)=f1(y-x2)|(d x)/(dy)|=ϵ^-(y-x2)
(e) none of these

Added by Martin F.

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