I am making some homework exercises at the moment and I was wondering if what I did in the following exercise was correct.


Solve $E(\sum_{k=0}^{N-1}(1-u_k)X_k + X_N) \rightarrow \max$, where $N\in\mathbb{N}$ is fixed $u_k$ are control variables in $[0,1]$, $X_k$ random variables with $x_0 = 1$ and $X_{k+1} = X_k + u_kX_k + Y_k+1$ and $Y_{k+1}$ are independent exponentially distributed random variables with mean $x_k$.


I first rewrote the objective function in the following way:

$$\sum_{k=0}^{N-1}E((1-u_k)X_k) + E(X_n) \rightarrow \max$$

Also, since Y is exponentially distributed $E(Y_{k+1}) = x_k$

As for the solution, I consider first the sum for $k = 1$ to maximize.

First I compute $X_1 = X_0 + u_0X_0 + Y_1$ where $E(Y_1) = 1$ since $x_0$ is $1$.

Also, $u_0 = 0$ since we want to maximize, and for $k=0$, $u_0$ is 1 for the maximum.

Hence, we get $E(1-u_1)X_1) \rightarrow \max = E((1-u_1)2)$ hence we get $u_1 = 0$

You do this for all $k$, is this the correct approach because I don't think I am correct..


Consider $N=2$, then your expression is $$E\{X_0-u_0X_0+X_1-u_1X_1+X_2\}.$$ Now substitute $X_1=X_0+u_0X_0+Y_0+1$ and $X_2=X_1+u_1X_1+Y_1+1$ and you will see the $-u_iX_i$, $i=\{0,1\}$ terms vanishes.$$E\{X_0+X_0+u_0X_0+Y_0+1+X_1+u_1X_1+Y_1+1\}.$$ Now substitute for $X_1$ again and you get $$E\{X_0+X_0+u_0X_0+Y_0+1+X_0+u_0X_0+Y_0+1+u_1(X_0+u_0X_0+Y_0+1)+Y_1+1\}.$$ Since $X_0=1$ and $EY_k$ is always positive for exponential random variables the controls $u_k$ should all be $1$ in order to maximize the expression.

  • $\begingroup$ Thank you for help, can you also show this for general N? $\endgroup$ – Nedellyzer Nov 6 '13 at 12:42
  • 1
    $\begingroup$ You can easily use mathematical induction to prove for general $N$. Since above we have proved for $N=2$, assume for $N=n$ all $u_0,\dots u_{n-1}$ are $1$ and show that for $N=n+1$ we need $u_n=1$ to maximize the expression. Which is easy to you taking $X_n$ and $X_n+1$ terms like we did for $N=2$. Then by induction we have for all $N$. $\endgroup$ – triomphe Nov 6 '13 at 15:25
  • $\begingroup$ Thanks but why is $X_1$ = $X_0 + u_0X_0 + Y_0 + 1$, I think it should be $X_0 + u_0X_0 + Y_1$? $\endgroup$ – Nedellyzer Nov 6 '13 at 15:53
  • $\begingroup$ But you define in your problem $X_{k+1}=X_k+u_kX_k+Y_k+1$ so for $X_1$ it is $Y_0$ then you should may be define $Y_{k+1}$. Is the $+1$ part of the index of $Y$? Either way the analysis doesn't change. $\endgroup$ – triomphe Nov 6 '13 at 20:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.