Quiz solutions for Math 103 Probability: Week 11

1. Events (B)

   (i) False. Let $\mathrm{\Omega }=\{1,2,3\}$, $A=\{1\}$, $B=\{2\}$.

   $A$ and $B$ are exclusive since $\{1\}\cap \{2\}=\mathrm{\varnothing }$ but $A^c=\{2,3\}$ and $B^c=\{1,3\}$ are not exclusive since $\{2,3\}\cap \{1,3\}=\{3\}$.

   (ii) False. Let $A$, $B$ be as above. Then $A\setminus B=\{1\}$, but $B\setminus A=\{2\}$.

   (iii) True

2. Vets (C)
   (i) violates axiom 2, (ii) is OK, (iii) violates axiom 1.

3. Conditional probabilities (A)

   (a) TRUE. $\mathrm{P}(A|B)=\frac{\mathrm{P}(A\cap B)}{\mathrm{P}(B)}=0$. Since $P(A\cap B)=0$ as $A$ and $B$ are exclusive events.

   (b) TRUE. If $A\subset B$ then $A\cap B=A$. So $\mathrm{P}(A|B)=\mathrm{P}(A\cap B)/\mathrm{P}(B)=\mathrm{P}(A)/\mathrm{P}(B)$.

   (c) TRUE. If $A\subset B$ then $A\cap B=A$. $\mathrm{P}(B|A)=\mathrm{P}(A\cap B)/\mathrm{P}(A)=\mathrm{P}(A)/\mathrm{P}(A)=1$.

4. Vectorised command (B)

   The code finds the maximum of each pair of elements from two vectors of the same length. For example:

   x <- c(1,2,3,4,5)

   y <- c(2,3,2,3,2)

   m <- rep(0,length(x))

   for (i in 1:length(x)) {

   m[i] <- max(x[i],y[i])

   }

   m

   2 3 3 4 5

   The same effect is given by the code in (B). (x >= y) returns a vector which is TRUE(=1) when the element of x is greater than the corresponding element of y and FALSE(=0) otherwise. So x*(x>=y) is a vector whose ith value is x[i] if x[i]>=y[i] and 0 otherwise. By a similar argument y*(x<y) is a vector whose ith value is y[i] if x[i]<y[i] and 0 otherwise. The result follows.

   Note that one might expect the code in (A) to give this result. However it returns a scalar rather than a vector, returning the maximum element of the concatenated vector c(x,y).

5. Recursion relationship II (A) The code should be modified to something like

   inhom_recursion <- function(A,B,k,u,v,N) {

   x <- c(u,v,rep( 1, N -2))

   for( n in 3:N){

   x[n] <- A*x[n-1] + B*x[n-2] + k

   }

   return(x)

   }

   We can then try out the different values of $k$ as follows:

   inhom_recursion(1.5,-1,0.3319,1,1/2,100)[100]

   -1.984991e-05

   inhom_recursion(1.5,-1,-0.1323,1,1/2,100)[100]

   -1.407286

   inhom_recursion(1.5,-1,1.006,1,1/2,100)[100]

   2.043578

   inhom_recursion(1.5,-1,-1.006,1,1/2,100)[100]

   -4.05599

   inhom_recursion(1.5,-1,0.4561,1,1/2,100)[100]

   0.3765042

   Hence the answer is A