Confusion about sum of random variables conditional probabilitiesLet X,Y be independent random variables, and Z = X + Y .. Then I want to calculate P r [ X = x ∣ Z = z ] .. My confusion is on evaluating this expression. On the one hand, I have P r [ X = x ∣ Z = z ] = P r [ Z − Y = x ∣ Z = z ] = P r [ z − Y = x ] = P r [ Y = z − x ] .But also, P r [ Z = z ∣ X = x ] = P r [ X + Y = z ∣ X = x ] = P r [ Y = z − x ] . So these two probabilities are equal? But P r [ Z = z ∣ X = x ] P r [ X = x ] = P r [ X = x ∣ Z = z ] P r [ Z = z ] and in general P r [ X = x ] and P r [ Z = z ] are not equal. I believe it should be P r [ X = x ∣ Z = z ] = P r [ Y = z − x ∣ Z = z ] but I don't think the conditional Z = z can be removed since Y and Z are not independent?I'm not sure whether the first equation holds either. For example, if I roll a fair six sided die X (numbered 1 to 6) and roll a fair ten sided die Y and take the sum, then P r [ X = 1 ∣ Z = 2 ] = 1 since the only possible outcome is ( x , y ) = ( 1 , 1 ) ,, and this is not equal to P r [ Y = ( 2 − 1 ) ] = 1 / 10.. On the other hand it is equal to P r [ Y = ( 2 − 1 ) ∣ Z = 2 ] = 1.. I think I'm making a mistake in one of these but it's not clear to me in which step.(The context of this was that X is a random variable with given distribution representing some unknown parameter and Y is a standard normal error. Then you observe z = x + y and want to estimate the X.)

Question

Confusion about sum of random variables conditional probabilitiesLet X,Y be independent random variables, and   Z  =  X  +  Y  .. Then I want to calculate   P  r  [  X  =  x  ∣  Z  =  z  ]  .. My confusion is on evaluating this expression. On the one hand, I have  P  r  [  X  =  x  ∣  Z  =  z  ]  =  P  r  [  Z  −  Y  =  x  ∣  Z  =  z  ]  =  P  r  [  z  −  Y  =  x  ]  =  P  r  [  Y  =  z  −  x  ]  .But also,   P  r  [  Z  =  z  ∣  X  =  x  ]  =  P  r  [  X  +  Y  =  z  ∣  X  =  x  ]  =  P  r  [  Y  =  z  −  x  ]  . So these two probabilities are equal? But   P  r  [  Z  =  z  ∣  X  =  x  ]  P  r  [  X  =  x  ]  =  P  r  [  X  =  x  ∣  Z  =  z  ]  P  r  [  Z  =  z  ] and in general   P  r  [  X  =  x  ] and   P  r  [  Z  =  z  ] are not equal. I believe it should be   P  r  [  X  =  x  ∣  Z  =  z  ]  =  P  r  [  Y  =  z  −  x  ∣  Z  =  z  ] but I don&#039;t think the conditional   Z  =  z can be removed since Y and Z are not independent?I&#039;m not sure whether the first equation holds either. For example, if I roll a fair six sided die X (numbered 1 to 6) and roll a fair ten sided die Y and take the sum, then   P  r  [  X  =  1  ∣  Z  =  2  ]  =  1 since the only possible outcome is   (  x  ,  y  )  =  (  1  ,  1  )  ,, and this is not equal to   P  r  [  Y  =  (  2  −  1  )  ]  =  1      /    10.. On the other hand it is equal to   P  r  [  Y  =  (  2  −  1  )  ∣  Z  =  2  ]  =  1.. I think I&#039;m making a mistake in one of these but it&#039;s not clear to me in which step.(The context of this was that X is a random variable with given distribution representing some unknown parameter and Y is a standard normal error. Then you observe   z  =  x  +  y and want to estimate the X.)

Ramiro Sosa · Accepted Answer

Step 1The mistake lies in this step:  P  r  [  Z  −  Y  =  x  ∣  Z  =  z  ]  =  P  r  [  z  −  Y  =  x  ]Note that   P  (  A      |    B  )  =            P      (      B              |            A      )      ⋅      P      (      A      )              P      (      B      )      . The actual evaluation is instead  P  (  Z  −  Y  =  x      |    Z  =  z  )  =            P      (      Z      =      z              |            Z      −      Y      =      x      )      ⋅      P      (      Z      −      Y      =      x      )              P      (      Z      =      z      )        =            P      (      Z      =      z              |            X      =      x      )      ⋅      P      (      X      =      x      )              P      (      Z      =      z      )      If we are given   X  =  x,   Z  =  z only when   Y  =  z  −  x. Thus,   P  (  Z  =  z      |    X  =  x  )  =  P  (  Y  =  z  −  x  ). The above formulation should make it clear that this is not true for   P  (  X  =  x      |    Z  =  z  )  . We now have  P  (  Z  −  Y  =  x      |    Z  =  z  )  =  P  (  Y  =  z  −  x  )  ⋅            P      (      X      =      x      )              P      (      Z      =      z      )      Step 2In your dice example, this evaluates as  P  (  X  =  1      |    Z  =  2  )  =  P  (  Y  =  1  )  ⋅            P      (      X      =      1      )              P      (      Z      =      2      )        =      1    10    ⋅            1      6              1      60        =  1 as expected.I shall take another example, to make it clearer. Let us calculate the probability   P  (  X  =  2      |    Z  =  5  ). We see that   Z  =  5 is achieved by the following pairs:  (  X  ,  Y  )  =  {  (  1  ,  4  )  ,  (  2  ,  3  )  ,  (  3  ,  2  )  ,  (  4  ,  1  )  }    ⟹    P  (  X  =  2      |    Z  =  5  )  =      1    4  Using the formula, we can calculate the same as  P  (  X  =  2      |    Z  =  5  )  =  P  (  Y  =  3  )  ⋅            P      (      X      =      2      )              P      (      Z      =      5      )        =      1    10    ⋅            1      6              4      60        =      1    4  which is the same as before.[  P  (  Z  =  5  )  =      4    60   as there are 60 possible (X,Y) and   Z  =  5 is only achieved by 4]

Let X,Y be independent random variables, and Z=X+Y.

Answered question

Answer & Explanation

New Questions in High school probability