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+2 votes
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Please solve the following question:

asked in Probability by Veteran (14.6k points)   | 202 views

2 Answers

+3 votes

Suppose you have an event that has 2-equally likely possibilities i.e, $X = a$ and $X = b$.

Clearly, $E(x)$ $=$ $(a+b)/2$ $=$ $1$

and $E(x^2)$ $=$ $(a^2 + b^2)/2$ $=$ $1$

Simplifying above two equations we get : $a + b = 2$ and $a^2 + b^2 = 2$

Putting $a = 2 - b$ on equation 2 and then solving $(2-b)^2 + b^2 = 2$, we get $a = 1$ and $b = 1$

Now,  E(X100) = (1100 + 1100)/2  = 1

Hence (B) is correct answer.

answered by Veteran (24.5k points)  
edited by
This question is based on probability distribution, you solved it like some apti question. E(x) is the mean.
can you please explain your logic?
0 votes

E(X100) = E( (X50)2 )

Now, as we go on reducing, we get E(X25). Now, E(X25) = E( X24.X)

We know that E(A.B) = E(A). E(B) for independent events A and B.

So, E( X24.X) = E(X24). E(X) = 1    (I am not sure at this step)

 

answered by Veteran (14.8k points)  
reshown by

E( X24.X) = E(X24). E(X) = 1    

this is correct when one function is independent of other i.e E(x.y) =E(X).E(Y).(when x,y are independent)

CAN WE SAY X24 is independent of X ?

@Agrasar. I think I got the solution. Going with the definition of expectation,

if s={x1,x2.........,xn} is the sample space then,

E(X2) = $\sum_{r\epsilon s}^{}$ X(r=s)2.P(r)     ..................(1)

E(X) = $\sum_{r\epsilon s}^{}$ X(r=s).P(r)        ..................(2)

Now, (1) and (2) are both equal which implies X(r=s)2=X(r=s)       ..................(3)

Now, lets consider what value we get for E(X3).

E(X3)

= $\sum_{r\epsilon s}^{}$ X(r=s)3.P(r)

= $\sum_{r\epsilon s}^{}$ X(r=s)2. X(r=s) .P(r)

= $\sum_{r\epsilon s}^{}$ X(r=s). X(r=s) .P(r)     .................from (3)

= E(X2)

=1

Similarly, we can go on computing for E(X100).

Thus, I think, E(X100) = 1

it should be 1

E(x100) =E(x50 ) (becoz E(X50^2) =E(X100)

            =E(X25)

            =E(X12 . X)

            =E(X6 . X)

            =E(X3.X)

           =E(X^4)

          =E((x^2)^2)

          =E(X^2)

          =E(X)

          =1
@cse23. They have give that E(X2)=E(X) and not E(X100)=E(X50). You are concluding that both things are same which is not correct I think unless they get proved by some formula.

So, both things are different. I had also done the same but later realised that its not correct.
but 100 =(50)^2 ryt?
@cse23. yes but there is difference between x and x^50


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