Probability calculations verification request: a distribution function related problem












0












$begingroup$


Let $X$ have distribution function $F(x)$ expressed by



begin{cases}
0 & mbox{if } x < 0 \
x/2 & mbox{if } 0leq x leq 2 \
1 & mbox{if } x geq 2
end{cases}



and let $Y = X^{2}$. Find



(a) $textbf{P}(frac{1}{2}leq Xleq frac{3}{2})$



(b) $textbf{P}(1leq X < 2)$



(c) $textbf{P}(Yleq X)$



(d) $textbf{P}(Xleq 2Y)$



(e) $textbf{P}(X+Yleqfrac{3}{4})$



(f) The distribution of $Z = sqrt{X}$.



MY ATTEMPT



(a) $displaystyletextbf{P}(0.5leq Xleq 1.5) = F(1.5) - lim_{xuparrow 0.5} F(x) = 0.75 - 0.25 = 0.5$



(b) $displaystyletextbf{P}(1leq X < 2) = lim_{xuparrow 2}F(x) - lim_{xuparrow 1}F(x) = 1 - 0.5 = 0.5$



(c) $displaystyletextbf{P}(Yleq X) = textbf{P}(X^{2} leq X) = textbf{P}(X(X-1) leq 0) = textbf{P}(0 leq Xleq 1)$



$displaystyletherefore textbf{P}(Yleq X) = F(1) - lim_{xuparrow 0}F(x) = 0.5 - 0 = 0.5$



(d) $displaystyletextbf{P}(X leq 2Y) = textbf{P}(Xleq 2X^{2}) = textbf{P}(X(2X-1)geq0) = 1 - textbf{P}(X<0.5)$



$displaystyletherefore textbf{P}(Xleq 2Y) = 1 - lim_{xuparrow 0.5}F(x) = 1 - 0 .25 = 0.75$



(e) $textbf{P}(X+Yleqfrac{3}{4}) = textbf{P}(4X^{2} + 4X - 3leq 0) = textbf{P}(-1.5 leq Xleq 0.5)$



$displaystyletherefore textbf{P}(X+Yleq 0.75) = F(0.5) - lim_{xuparrow-1.5}F(x) = 0.25 - 0 = 0.25$



(f) I am not sure how to solve it. Here is my attempt



begin{align*}
textbf{P}(Zleq z) = textbf{P}(sqrt{X}leq z) = textbf{P}(Xleq z^{2}) = F_{X}(z^{2})
end{align*}



More precisely, the distribution function of $Z$ is given by



begin{cases}
0 & mbox{if } z < 0 \
z^{2}/2 & mbox{if } 0leq z leq sqrt{2} \
1 & mbox{if } z geqsqrt{2}
end{cases}



Could someone please double-check my results? I thank in advance for any contribution.



EDIT



I have edited the answer according to Kavi's contribution. Any other contribution is equally welcome as well.






probability probability-theory proof-verification probability-distributions







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Let $X$ have distribution function $F(x)$ expressed by



    begin{cases}
    0 & mbox{if } x < 0 \
    x/2 & mbox{if } 0leq x leq 2 \
    1 & mbox{if } x geq 2
    end{cases}



    and let $Y = X^{2}$. Find



    (a) $textbf{P}(frac{1}{2}leq Xleq frac{3}{2})$



    (b) $textbf{P}(1leq X < 2)$



    (c) $textbf{P}(Yleq X)$



    (d) $textbf{P}(Xleq 2Y)$



    (e) $textbf{P}(X+Yleqfrac{3}{4})$



    (f) The distribution of $Z = sqrt{X}$.



    MY ATTEMPT



    (a) $displaystyletextbf{P}(0.5leq Xleq 1.5) = F(1.5) - lim_{xuparrow 0.5} F(x) = 0.75 - 0.25 = 0.5$



    (b) $displaystyletextbf{P}(1leq X < 2) = lim_{xuparrow 2}F(x) - lim_{xuparrow 1}F(x) = 1 - 0.5 = 0.5$



    (c) $displaystyletextbf{P}(Yleq X) = textbf{P}(X^{2} leq X) = textbf{P}(X(X-1) leq 0) = textbf{P}(0 leq Xleq 1)$



    $displaystyletherefore textbf{P}(Yleq X) = F(1) - lim_{xuparrow 0}F(x) = 0.5 - 0 = 0.5$



    (d) $displaystyletextbf{P}(X leq 2Y) = textbf{P}(Xleq 2X^{2}) = textbf{P}(X(2X-1)geq0) = 1 - textbf{P}(X<0.5)$



    $displaystyletherefore textbf{P}(Xleq 2Y) = 1 - lim_{xuparrow 0.5}F(x) = 1 - 0 .25 = 0.75$



    (e) $textbf{P}(X+Yleqfrac{3}{4}) = textbf{P}(4X^{2} + 4X - 3leq 0) = textbf{P}(-1.5 leq Xleq 0.5)$



    $displaystyletherefore textbf{P}(X+Yleq 0.75) = F(0.5) - lim_{xuparrow-1.5}F(x) = 0.25 - 0 = 0.25$



    (f) I am not sure how to solve it. Here is my attempt



    begin{align*}
    textbf{P}(Zleq z) = textbf{P}(sqrt{X}leq z) = textbf{P}(Xleq z^{2}) = F_{X}(z^{2})
    end{align*}



    More precisely, the distribution function of $Z$ is given by



    begin{cases}
    0 & mbox{if } z < 0 \
    z^{2}/2 & mbox{if } 0leq z leq sqrt{2} \
    1 & mbox{if } z geqsqrt{2}
    end{cases}



    Could someone please double-check my results? I thank in advance for any contribution.



    EDIT



    I have edited the answer according to Kavi's contribution. Any other contribution is equally welcome as well.






    probability probability-theory proof-verification probability-distributions







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $X$ have distribution function $F(x)$ expressed by



      begin{cases}
      0 & mbox{if } x < 0 \
      x/2 & mbox{if } 0leq x leq 2 \
      1 & mbox{if } x geq 2
      end{cases}



      and let $Y = X^{2}$. Find



      (a) $textbf{P}(frac{1}{2}leq Xleq frac{3}{2})$



      (b) $textbf{P}(1leq X < 2)$



      (c) $textbf{P}(Yleq X)$



      (d) $textbf{P}(Xleq 2Y)$



      (e) $textbf{P}(X+Yleqfrac{3}{4})$



      (f) The distribution of $Z = sqrt{X}$.



      MY ATTEMPT



      (a) $displaystyletextbf{P}(0.5leq Xleq 1.5) = F(1.5) - lim_{xuparrow 0.5} F(x) = 0.75 - 0.25 = 0.5$



      (b) $displaystyletextbf{P}(1leq X < 2) = lim_{xuparrow 2}F(x) - lim_{xuparrow 1}F(x) = 1 - 0.5 = 0.5$



      (c) $displaystyletextbf{P}(Yleq X) = textbf{P}(X^{2} leq X) = textbf{P}(X(X-1) leq 0) = textbf{P}(0 leq Xleq 1)$



      $displaystyletherefore textbf{P}(Yleq X) = F(1) - lim_{xuparrow 0}F(x) = 0.5 - 0 = 0.5$



      (d) $displaystyletextbf{P}(X leq 2Y) = textbf{P}(Xleq 2X^{2}) = textbf{P}(X(2X-1)geq0) = 1 - textbf{P}(X<0.5)$



      $displaystyletherefore textbf{P}(Xleq 2Y) = 1 - lim_{xuparrow 0.5}F(x) = 1 - 0 .25 = 0.75$



      (e) $textbf{P}(X+Yleqfrac{3}{4}) = textbf{P}(4X^{2} + 4X - 3leq 0) = textbf{P}(-1.5 leq Xleq 0.5)$



      $displaystyletherefore textbf{P}(X+Yleq 0.75) = F(0.5) - lim_{xuparrow-1.5}F(x) = 0.25 - 0 = 0.25$



      (f) I am not sure how to solve it. Here is my attempt



      begin{align*}
      textbf{P}(Zleq z) = textbf{P}(sqrt{X}leq z) = textbf{P}(Xleq z^{2}) = F_{X}(z^{2})
      end{align*}



      More precisely, the distribution function of $Z$ is given by



      begin{cases}
      0 & mbox{if } z < 0 \
      z^{2}/2 & mbox{if } 0leq z leq sqrt{2} \
      1 & mbox{if } z geqsqrt{2}
      end{cases}



      Could someone please double-check my results? I thank in advance for any contribution.



      EDIT



      I have edited the answer according to Kavi's contribution. Any other contribution is equally welcome as well.






      probability probability-theory proof-verification probability-distributions







      share|cite|improve this question











      $endgroup$




      Let $X$ have distribution function $F(x)$ expressed by



      begin{cases}
      0 & mbox{if } x < 0 \
      x/2 & mbox{if } 0leq x leq 2 \
      1 & mbox{if } x geq 2
      end{cases}



      and let $Y = X^{2}$. Find



      (a) $textbf{P}(frac{1}{2}leq Xleq frac{3}{2})$



      (b) $textbf{P}(1leq X < 2)$



      (c) $textbf{P}(Yleq X)$



      (d) $textbf{P}(Xleq 2Y)$



      (e) $textbf{P}(X+Yleqfrac{3}{4})$



      (f) The distribution of $Z = sqrt{X}$.



      MY ATTEMPT



      (a) $displaystyletextbf{P}(0.5leq Xleq 1.5) = F(1.5) - lim_{xuparrow 0.5} F(x) = 0.75 - 0.25 = 0.5$



      (b) $displaystyletextbf{P}(1leq X < 2) = lim_{xuparrow 2}F(x) - lim_{xuparrow 1}F(x) = 1 - 0.5 = 0.5$



      (c) $displaystyletextbf{P}(Yleq X) = textbf{P}(X^{2} leq X) = textbf{P}(X(X-1) leq 0) = textbf{P}(0 leq Xleq 1)$



      $displaystyletherefore textbf{P}(Yleq X) = F(1) - lim_{xuparrow 0}F(x) = 0.5 - 0 = 0.5$



      (d) $displaystyletextbf{P}(X leq 2Y) = textbf{P}(Xleq 2X^{2}) = textbf{P}(X(2X-1)geq0) = 1 - textbf{P}(X<0.5)$



      $displaystyletherefore textbf{P}(Xleq 2Y) = 1 - lim_{xuparrow 0.5}F(x) = 1 - 0 .25 = 0.75$



      (e) $textbf{P}(X+Yleqfrac{3}{4}) = textbf{P}(4X^{2} + 4X - 3leq 0) = textbf{P}(-1.5 leq Xleq 0.5)$



      $displaystyletherefore textbf{P}(X+Yleq 0.75) = F(0.5) - lim_{xuparrow-1.5}F(x) = 0.25 - 0 = 0.25$



      (f) I am not sure how to solve it. Here is my attempt



      begin{align*}
      textbf{P}(Zleq z) = textbf{P}(sqrt{X}leq z) = textbf{P}(Xleq z^{2}) = F_{X}(z^{2})
      end{align*}



      More precisely, the distribution function of $Z$ is given by



      begin{cases}
      0 & mbox{if } z < 0 \
      z^{2}/2 & mbox{if } 0leq z leq sqrt{2} \
      1 & mbox{if } z geqsqrt{2}
      end{cases}



      Could someone please double-check my results? I thank in advance for any contribution.



      EDIT



      I have edited the answer according to Kavi's contribution. Any other contribution is equally welcome as well.






      probability probability-theory proof-verification probability-distributions




      probability probability-theory proof-verification probability-distributions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 22 at 19:52







      user1337

















      asked Jan 21 at 23:14









      user1337user1337

      46110




      46110






















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          $begingroup$

          I think there is a mistake in e). $4X^{2}+4X-3 leq 0$ gives $-frac 3 2 leq X leq frac 1 2$. In f) you should replace the condition $0 leq z leq 2$ by $0 leq z^{2} leq 2$ or $0 leq z leq sqrt 2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks, Kavi, for the contribution. I have edited the question according to your observations.
            $endgroup$
            – user1337
            Jan 21 at 23:51











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          $begingroup$

          I think there is a mistake in e). $4X^{2}+4X-3 leq 0$ gives $-frac 3 2 leq X leq frac 1 2$. In f) you should replace the condition $0 leq z leq 2$ by $0 leq z^{2} leq 2$ or $0 leq z leq sqrt 2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks, Kavi, for the contribution. I have edited the question according to your observations.
            $endgroup$
            – user1337
            Jan 21 at 23:51
















          1












          $begingroup$

          I think there is a mistake in e). $4X^{2}+4X-3 leq 0$ gives $-frac 3 2 leq X leq frac 1 2$. In f) you should replace the condition $0 leq z leq 2$ by $0 leq z^{2} leq 2$ or $0 leq z leq sqrt 2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks, Kavi, for the contribution. I have edited the question according to your observations.
            $endgroup$
            – user1337
            Jan 21 at 23:51














          1












          1








          1





          $begingroup$

          I think there is a mistake in e). $4X^{2}+4X-3 leq 0$ gives $-frac 3 2 leq X leq frac 1 2$. In f) you should replace the condition $0 leq z leq 2$ by $0 leq z^{2} leq 2$ or $0 leq z leq sqrt 2$.






          share|cite|improve this answer









          $endgroup$



          I think there is a mistake in e). $4X^{2}+4X-3 leq 0$ gives $-frac 3 2 leq X leq frac 1 2$. In f) you should replace the condition $0 leq z leq 2$ by $0 leq z^{2} leq 2$ or $0 leq z leq sqrt 2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 21 at 23:40









          Kavi Rama MurthyKavi Rama Murthy

          62.7k42262




          62.7k42262












          • $begingroup$
            Thanks, Kavi, for the contribution. I have edited the question according to your observations.
            $endgroup$
            – user1337
            Jan 21 at 23:51


















          • $begingroup$
            Thanks, Kavi, for the contribution. I have edited the question according to your observations.
            $endgroup$
            – user1337
            Jan 21 at 23:51
















          $begingroup$
          Thanks, Kavi, for the contribution. I have edited the question according to your observations.
          $endgroup$
          – user1337
          Jan 21 at 23:51




          $begingroup$
          Thanks, Kavi, for the contribution. I have edited the question according to your observations.
          $endgroup$
          – user1337
          Jan 21 at 23:51


















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