Expected value

Question

in C language rand() returns a random integer number.  Following is a function definition.

int f() {
     return (rand() & 1) ? 1:0;
}

Then what will be the expected value of count ?

 count = 0;
 for(i=1;i<=n;i++) {
    if(f() && f()) {
       count++;
    }
 }

& is bitwise AND

&& is logical AND

and assume required seed has been initialized.

Comments

is it $n/4$?

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yes...

Answer 1

The function $f()$ is checking if the random integer generated by $rand()$ is odd. The expression $n&1$ is used for checking the parity of $n$ (odd or even), so $f()$ returns $1$ if the random number is odd and return $0$ otherwise.

Now inside the for loop, the variable $count$ is incremented only if both the $f()$'s inside the if condition are true. This happens when $2$ odd random numbers are generated.

For calculating the expected value of $count$, let us have an indicator random variable $X_i$, such that $X_i = 1$ if $count$ is incremented at the $ith$ iteration in the loop and $X_i = 0$ otherwise. Then $E[Xi] = P(Xi) = 1/4$ (Odd Even, Even Odd, Odd Odd, Even Even There are 4 possibilities. Favorable case is Odd Odd) The expected value of $count$ call this $E[Z]$ is given by $E[Z] = E[X1] + E[X2] + E[X3] + ... + E[XN]$. So $E[Z] = N * 1/4 = N/4$