Solving $P(2<X^2<5)$ given PDF $f(x)$ and CDF $F(x)$ of X

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
1
down vote

favorite












If it was $P(2<X<5)$ I would just use $F(5)-F(2)$.



For this case do I solve $2$ cases with



$P(sqrt2<X<sqrt5)$ and $P(-sqrt 5 < X < - sqrt2)$



My random variable is defined for all real numbers so I think this is valid, is this correct? Note my pdf has an absolute value so the CDF is defined separately for $xle0$ and $x>0$ which also supports doing that.




probability




share|cite|improve this question


























    up vote
    1
    down vote

    favorite












    If it was $P(2<X<5)$ I would just use $F(5)-F(2)$.



    For this case do I solve $2$ cases with



    $P(sqrt2<X<sqrt5)$ and $P(-sqrt 5 < X < - sqrt2)$



    My random variable is defined for all real numbers so I think this is valid, is this correct? Note my pdf has an absolute value so the CDF is defined separately for $xle0$ and $x>0$ which also supports doing that.




    probability




    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      If it was $P(2<X<5)$ I would just use $F(5)-F(2)$.



      For this case do I solve $2$ cases with



      $P(sqrt2<X<sqrt5)$ and $P(-sqrt 5 < X < - sqrt2)$



      My random variable is defined for all real numbers so I think this is valid, is this correct? Note my pdf has an absolute value so the CDF is defined separately for $xle0$ and $x>0$ which also supports doing that.




      probability




      share|cite|improve this question














      If it was $P(2<X<5)$ I would just use $F(5)-F(2)$.



      For this case do I solve $2$ cases with



      $P(sqrt2<X<sqrt5)$ and $P(-sqrt 5 < X < - sqrt2)$



      My random variable is defined for all real numbers so I think this is valid, is this correct? Note my pdf has an absolute value so the CDF is defined separately for $xle0$ and $x>0$ which also supports doing that.





      probability





      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 26 at 22:22







      user550230

















      asked Aug 26 at 21:54









      glockm15

      1649




      1649




















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          You have the right idea but it is important to note that $$-sqrt5 < -sqrt2,$$ thus your second probability expression should read $$Pr[-sqrt5 < X < -sqrt2],$$ not the other way around--which would give zero.






          share|cite|improve this answer




















          • So when I have found the probabilities for each case would it make more sense to add probability of the two answers.As x being positive or negative are independent events. Or just report each probability found for each case separately.
            – glockm15
            Aug 26 at 22:08







          • 1




            You do need to add the probabilities together; that is to say, $$Pr[2 < X^2 < 5] = Pr[-sqrt5 < X < -sqrt2] + Pr[sqrt2 < X < sqrt5].$$ Note that this works because the intervals over which you are evaluating those probabilities are disjoint.
            – heropup
            Aug 26 at 22:17










          Your Answer




          StackExchange.ifUsing("editor", function ()
          return StackExchange.using("mathjaxEditing", function ()
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          );
          );
          , "mathjax-editing");

          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "69"
          ;
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function()
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled)
          StackExchange.using("snippets", function()
          createEditor();
          );

          else
          createEditor();

          );

          function createEditor()
          StackExchange.prepareEditor(
          heartbeatType: 'answer',
          convertImagesToLinks: true,
          noModals: false,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );













           

          draft saved


          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2895569%2fsolving-p2x25-given-pdf-fx-and-cdf-fx-of-x%23new-answer', 'question_page');

          );

          Post as a guest

















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          You have the right idea but it is important to note that $$-sqrt5 < -sqrt2,$$ thus your second probability expression should read $$Pr[-sqrt5 < X < -sqrt2],$$ not the other way around--which would give zero.






          share|cite|improve this answer




















          • So when I have found the probabilities for each case would it make more sense to add probability of the two answers.As x being positive or negative are independent events. Or just report each probability found for each case separately.
            – glockm15
            Aug 26 at 22:08







          • 1




            You do need to add the probabilities together; that is to say, $$Pr[2 < X^2 < 5] = Pr[-sqrt5 < X < -sqrt2] + Pr[sqrt2 < X < sqrt5].$$ Note that this works because the intervals over which you are evaluating those probabilities are disjoint.
            – heropup
            Aug 26 at 22:17














          up vote
          1
          down vote



          accepted










          You have the right idea but it is important to note that $$-sqrt5 < -sqrt2,$$ thus your second probability expression should read $$Pr[-sqrt5 < X < -sqrt2],$$ not the other way around--which would give zero.






          share|cite|improve this answer




















          • So when I have found the probabilities for each case would it make more sense to add probability of the two answers.As x being positive or negative are independent events. Or just report each probability found for each case separately.
            – glockm15
            Aug 26 at 22:08







          • 1




            You do need to add the probabilities together; that is to say, $$Pr[2 < X^2 < 5] = Pr[-sqrt5 < X < -sqrt2] + Pr[sqrt2 < X < sqrt5].$$ Note that this works because the intervals over which you are evaluating those probabilities are disjoint.
            – heropup
            Aug 26 at 22:17












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          You have the right idea but it is important to note that $$-sqrt5 < -sqrt2,$$ thus your second probability expression should read $$Pr[-sqrt5 < X < -sqrt2],$$ not the other way around--which would give zero.






          share|cite|improve this answer












          You have the right idea but it is important to note that $$-sqrt5 < -sqrt2,$$ thus your second probability expression should read $$Pr[-sqrt5 < X < -sqrt2],$$ not the other way around--which would give zero.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 26 at 22:01









          heropup

          60.3k65895




          60.3k65895











          • So when I have found the probabilities for each case would it make more sense to add probability of the two answers.As x being positive or negative are independent events. Or just report each probability found for each case separately.
            – glockm15
            Aug 26 at 22:08







          • 1




            You do need to add the probabilities together; that is to say, $$Pr[2 < X^2 < 5] = Pr[-sqrt5 < X < -sqrt2] + Pr[sqrt2 < X < sqrt5].$$ Note that this works because the intervals over which you are evaluating those probabilities are disjoint.
            – heropup
            Aug 26 at 22:17
















          • So when I have found the probabilities for each case would it make more sense to add probability of the two answers.As x being positive or negative are independent events. Or just report each probability found for each case separately.
            – glockm15
            Aug 26 at 22:08







          • 1




            You do need to add the probabilities together; that is to say, $$Pr[2 < X^2 < 5] = Pr[-sqrt5 < X < -sqrt2] + Pr[sqrt2 < X < sqrt5].$$ Note that this works because the intervals over which you are evaluating those probabilities are disjoint.
            – heropup
            Aug 26 at 22:17















          So when I have found the probabilities for each case would it make more sense to add probability of the two answers.As x being positive or negative are independent events. Or just report each probability found for each case separately.
          – glockm15
          Aug 26 at 22:08





          So when I have found the probabilities for each case would it make more sense to add probability of the two answers.As x being positive or negative are independent events. Or just report each probability found for each case separately.
          – glockm15
          Aug 26 at 22:08





          1




          1




          You do need to add the probabilities together; that is to say, $$Pr[2 < X^2 < 5] = Pr[-sqrt5 < X < -sqrt2] + Pr[sqrt2 < X < sqrt5].$$ Note that this works because the intervals over which you are evaluating those probabilities are disjoint.
          – heropup
          Aug 26 at 22:17




          You do need to add the probabilities together; that is to say, $$Pr[2 < X^2 < 5] = Pr[-sqrt5 < X < -sqrt2] + Pr[sqrt2 < X < sqrt5].$$ Note that this works because the intervals over which you are evaluating those probabilities are disjoint.
          – heropup
          Aug 26 at 22:17

















           

          draft saved


          draft discarded















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2895569%2fsolving-p2x25-given-pdf-fx-and-cdf-fx-of-x%23new-answer', 'question_page');

          );

          Post as a guest
































































          -

          這個網誌中的熱門文章

          Is there any way to eliminate the singular point to solve this integral by hand or by approximations?

          圖片
          Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite 1 $$int_0^1 (1+(x^-p-1)^1-p)^frac1pdx, pin mathbbR, pge 1$$ I find that $x=0,1$ are singular points(discontinuous points of this integral), is there a way to eliminate those singular points to make sure that this integral can be calculated by hand or by Taylor approximations. In either cases, what are those steps of calculating the integral? calculus real-analysis analysis approximation share | cite | improve this question edited Sep 11 at 5:07 asked Sep 11 at 4:48 Mclalalala 126 8 1 What is $p$? Please give more information to make it complete. – xbh Sep 11 at 5:02 @xbh already fixed it, thanks for reminding. – Mclalalala Sep 11 at 5:10 Seems that $0$ is not singular here… [I might compute wrong] – xbh Sep 11 at 5:30 @xbh can you show me how to get the close formula for the integral above? – Mclalalala Sep 11 at 5:33
          閱讀完整內容

          Why am i infinitely getting the same tweet with the Twitter Search API?

          圖片
          up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia
          閱讀完整內容

          Carbon dioxide

          圖片
          "CO2" redirects here. For other uses, see CO2 (disambiguation). Carbon dioxide Names Other names Carbonic acid gas Carbonic anhydride Carbonic oxide Carbon oxide Carbon(IV) oxide Dry ice (solid phase) Identifiers CAS Number 124-38-9   Y 3D model (JSmol) Interactive image Interactive image 3DMet B01131 Beilstein Reference 1900390 ChEBI CHEBI:16526   Y ChEMBL ChEMBL1231871   N ChemSpider 274   Y ECHA InfoCard 100.004.271 EC Number 204-696-9 E number E290 (preservatives) Gmelin Reference 989 KEGG D00004   Y MeSH Carbon+dioxide PubChem CID 280 RTECS number FF6400000 UNII 142M471B3J   Y UN number 1013 (gas), 1845 (solid) InChI InChI=1S/CO2/c2-1-3   Y Key: CURLTUGMZLYLDI-UHFFFAOYSA-N   Y InChI=1/CO2/c2-1-3 Key: CURLTUGMZLYLDI-UHFFFAOYAO SMILES O=C=O C(=O)=O Properties Chemical formula C O 2 Molar mass 44.01 g·mol −1 Appearance Colorless gas Odor Low concentrations: none High concentrations: sharp; acidic [1] Density 1562 kg/m 3 (solid at 1 atm and −78.5 °C) 1101 kg/m 3 (liquid at
          閱讀完整內容