Find $P(X=Y+1)$











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Suppose that X and Y are integer valued random variables with joint probability mass function given by
$p_{X,Y}(a, b)=begin{cases} frac{1}{4a}, & 1leq bleq aleq 4\ 0, & text{otherwise}. end{cases}.$



(b) Find the marginal probability mass functions of X and Y.
(c) Find $P(X=Y+1)$



Since this is a discrete random variable, I want to construct a table of joint distribution to find the sum of rows, columns for pmf of X and Y, but I'm having trouble with it. Anyone can help with part (b) and (c)?






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    up vote
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    down vote

    favorite












    Suppose that X and Y are integer valued random variables with joint probability mass function given by
    $p_{X,Y}(a, b)=begin{cases} frac{1}{4a}, & 1leq bleq aleq 4\ 0, & text{otherwise}. end{cases}.$



    (b) Find the marginal probability mass functions of X and Y.
    (c) Find $P(X=Y+1)$



    Since this is a discrete random variable, I want to construct a table of joint distribution to find the sum of rows, columns for pmf of X and Y, but I'm having trouble with it. Anyone can help with part (b) and (c)?






    probability







    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Suppose that X and Y are integer valued random variables with joint probability mass function given by
      $p_{X,Y}(a, b)=begin{cases} frac{1}{4a}, & 1leq bleq aleq 4\ 0, & text{otherwise}. end{cases}.$



      (b) Find the marginal probability mass functions of X and Y.
      (c) Find $P(X=Y+1)$



      Since this is a discrete random variable, I want to construct a table of joint distribution to find the sum of rows, columns for pmf of X and Y, but I'm having trouble with it. Anyone can help with part (b) and (c)?






      probability







      share|cite|improve this question













      Suppose that X and Y are integer valued random variables with joint probability mass function given by
      $p_{X,Y}(a, b)=begin{cases} frac{1}{4a}, & 1leq bleq aleq 4\ 0, & text{otherwise}. end{cases}.$



      (b) Find the marginal probability mass functions of X and Y.
      (c) Find $P(X=Y+1)$



      Since this is a discrete random variable, I want to construct a table of joint distribution to find the sum of rows, columns for pmf of X and Y, but I'm having trouble with it. Anyone can help with part (b) and (c)?






      probability




      probability






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      asked Nov 14 at 4:18









      dxdydz

      989




      989






















          1 Answer
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          The table is not hard.   Each cell in 4 rows ($1leq aleq 4$) of $a$ columns ($1leq bleq a$), contains $tfrac 1{4a}$, the rest of the cells are zero .



          $$p_{X,Y}(a,b)=tfrac 1{4a}mathbf 1_{1leq bleq aleq 4}\boxed{begin{array}{c|c:c:c:c|c}abackslash b & 1 & 2 & 3 & 4 &\hline 1 & tfrac 14 &0&0&0\ hdashline 2 & tfrac 18&tfrac 18&0&0\ hdashline 3 &tfrac 1{12}&tfrac 1{12}&tfrac 1{12}&0\ hdashline 4 & tfrac 1{16}&tfrac 1{16}&tfrac 1{16}&tfrac 1{16}\hline ~&&&&&1end{array}}$$






          share|cite|improve this answer



















          • 1




            So $P(X=Y+1) = P(X=2, Y=1) + P(X=3, Y=2) + P(X=4, Y=3) = 1/8 + 1/12 +1/16 = 13/48$?
            – dxdydz
            Nov 14 at 5:08










          • Indeed. @dxdydz
            – Graham Kemp
            Nov 14 at 11:59











          Your Answer





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          The table is not hard.   Each cell in 4 rows ($1leq aleq 4$) of $a$ columns ($1leq bleq a$), contains $tfrac 1{4a}$, the rest of the cells are zero .



          $$p_{X,Y}(a,b)=tfrac 1{4a}mathbf 1_{1leq bleq aleq 4}\boxed{begin{array}{c|c:c:c:c|c}abackslash b & 1 & 2 & 3 & 4 &\hline 1 & tfrac 14 &0&0&0\ hdashline 2 & tfrac 18&tfrac 18&0&0\ hdashline 3 &tfrac 1{12}&tfrac 1{12}&tfrac 1{12}&0\ hdashline 4 & tfrac 1{16}&tfrac 1{16}&tfrac 1{16}&tfrac 1{16}\hline ~&&&&&1end{array}}$$






          share|cite|improve this answer



















          • 1




            So $P(X=Y+1) = P(X=2, Y=1) + P(X=3, Y=2) + P(X=4, Y=3) = 1/8 + 1/12 +1/16 = 13/48$?
            – dxdydz
            Nov 14 at 5:08










          • Indeed. @dxdydz
            – Graham Kemp
            Nov 14 at 11:59















          up vote
          1
          down vote



          accepted










          The table is not hard.   Each cell in 4 rows ($1leq aleq 4$) of $a$ columns ($1leq bleq a$), contains $tfrac 1{4a}$, the rest of the cells are zero .



          $$p_{X,Y}(a,b)=tfrac 1{4a}mathbf 1_{1leq bleq aleq 4}\boxed{begin{array}{c|c:c:c:c|c}abackslash b & 1 & 2 & 3 & 4 &\hline 1 & tfrac 14 &0&0&0\ hdashline 2 & tfrac 18&tfrac 18&0&0\ hdashline 3 &tfrac 1{12}&tfrac 1{12}&tfrac 1{12}&0\ hdashline 4 & tfrac 1{16}&tfrac 1{16}&tfrac 1{16}&tfrac 1{16}\hline ~&&&&&1end{array}}$$






          share|cite|improve this answer



















          • 1




            So $P(X=Y+1) = P(X=2, Y=1) + P(X=3, Y=2) + P(X=4, Y=3) = 1/8 + 1/12 +1/16 = 13/48$?
            – dxdydz
            Nov 14 at 5:08










          • Indeed. @dxdydz
            – Graham Kemp
            Nov 14 at 11:59













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          The table is not hard.   Each cell in 4 rows ($1leq aleq 4$) of $a$ columns ($1leq bleq a$), contains $tfrac 1{4a}$, the rest of the cells are zero .



          $$p_{X,Y}(a,b)=tfrac 1{4a}mathbf 1_{1leq bleq aleq 4}\boxed{begin{array}{c|c:c:c:c|c}abackslash b & 1 & 2 & 3 & 4 &\hline 1 & tfrac 14 &0&0&0\ hdashline 2 & tfrac 18&tfrac 18&0&0\ hdashline 3 &tfrac 1{12}&tfrac 1{12}&tfrac 1{12}&0\ hdashline 4 & tfrac 1{16}&tfrac 1{16}&tfrac 1{16}&tfrac 1{16}\hline ~&&&&&1end{array}}$$






          share|cite|improve this answer














          The table is not hard.   Each cell in 4 rows ($1leq aleq 4$) of $a$ columns ($1leq bleq a$), contains $tfrac 1{4a}$, the rest of the cells are zero .



          $$p_{X,Y}(a,b)=tfrac 1{4a}mathbf 1_{1leq bleq aleq 4}\boxed{begin{array}{c|c:c:c:c|c}abackslash b & 1 & 2 & 3 & 4 &\hline 1 & tfrac 14 &0&0&0\ hdashline 2 & tfrac 18&tfrac 18&0&0\ hdashline 3 &tfrac 1{12}&tfrac 1{12}&tfrac 1{12}&0\ hdashline 4 & tfrac 1{16}&tfrac 1{16}&tfrac 1{16}&tfrac 1{16}\hline ~&&&&&1end{array}}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 14 at 5:06

























          answered Nov 14 at 5:00









          Graham Kemp

          84.2k43378




          84.2k43378








          • 1




            So $P(X=Y+1) = P(X=2, Y=1) + P(X=3, Y=2) + P(X=4, Y=3) = 1/8 + 1/12 +1/16 = 13/48$?
            – dxdydz
            Nov 14 at 5:08










          • Indeed. @dxdydz
            – Graham Kemp
            Nov 14 at 11:59














          • 1




            So $P(X=Y+1) = P(X=2, Y=1) + P(X=3, Y=2) + P(X=4, Y=3) = 1/8 + 1/12 +1/16 = 13/48$?
            – dxdydz
            Nov 14 at 5:08










          • Indeed. @dxdydz
            – Graham Kemp
            Nov 14 at 11:59








          1




          1




          So $P(X=Y+1) = P(X=2, Y=1) + P(X=3, Y=2) + P(X=4, Y=3) = 1/8 + 1/12 +1/16 = 13/48$?
          – dxdydz
          Nov 14 at 5:08




          So $P(X=Y+1) = P(X=2, Y=1) + P(X=3, Y=2) + P(X=4, Y=3) = 1/8 + 1/12 +1/16 = 13/48$?
          – dxdydz
          Nov 14 at 5:08












          Indeed. @dxdydz
          – Graham Kemp
          Nov 14 at 11:59




          Indeed. @dxdydz
          – Graham Kemp
          Nov 14 at 11:59


















           

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