Pure mathematician needs help with probability

[gccdman]

Hi,

I've posted this on other forums and have had little to no help. I thought this forum looked very good so hoping someone can help.

Sorry for the non specific title of the thread, I don't know what the following problem should be called! I don't think it is too hard but I stopped studying probability after my first year at uni and do not have much intuition in this field of mathematics. Anyway, here goes...

If X₁ and X₂ are random variables that can take any values in the finite set {0,2,4,...,2n} all with equal chance then how do I work out the probability, given an arbitrary positive real a that:

X₁ <= a <= X₁+X₂

If we call this probability P(a), how do we find a such that

P(a) => P(b) for arbitrary positive real b

Acting on advice I have calculated the pmf for X₁ and Y= X₁ + X₂

X₁: S -> R is defined by th identity function X₁(s) = s.

Hence, f₁(x) = P(X₁ = x) = P({s € S: X₁(s) = x}) = P({s€S: s = x) = 0 if x does not belong to {0,2,...,2n}; 1/(n+1) if x € S

Since X₂: S -> R is defined by X₂(s) = s we have f₁(x) = f₂(x).

Let Y = X₁ + X₂. Y: SxS -> R is defined by Y(s,s') = X₁(s) + X₂(s') = s + s'.

Hence f_Y(x) = P(Y = x) = P({(s,s') € SxS : Y(s,s') = x}) = P({(s,s') € SxS: s + s' = x) = 0 if x does not belong to {0,2,...,4n}; (x+2)/2(n+1)² if x € {0,2,...,2n}; (2(2n+1)-x)/2(n+1)² if x € {2n, 2n+2,..., 4n}

However, after this no further advice followed.

In the process of trying to solve it my self and guessing what the further advice may have been I have calculated explicitly the cdf for X₁ and Y = X₁ + X₂ which I shall denote F_X1(x) and F_Y(x) respectively.

Let E be the event X₁ <= x, E' the event x <= Y.

Now if E and E' were independant by definiton P(E and E') = P(E)P(E') = F_X1(x)(1-F_Y(x)+f_Y(x)) and hence the problem would be solved. However, E' is conditional on E. I would go on to solve this problem by working out f_Y/E(x) and F_Y/E(x),

for then P(E and E') = P(E)P(E'/E) = F_X1(x)(1-F_Y/E(x)+f_Y/E(x)).

Is this the right way of solving the problem? Is there a way of calculating f_Y/E(x) and F_Y/E(x) from f_X1(x), F_X1(x), f_Y(x) and F_Y(x)?

Any advice would be much appeciated,

Thanks

 

[lanedance]

hi, not exactly an explicit expression & my probabilty's a little sketchy, (enough disclaimers) this is what I came up with

first the probabilty for the single random variable is:

[tex]P(X=m) = \frac{1}{(n+1)} [/tex]
for m in {0,2,...2n}, 0 otherwise

the probability of Y = X1 + X2 will then be a convolution of X with itself, where the sum is over all values of r
[tex] P(Y=(X_1+X_2)=m) = \sum_r P(X_1=r)P(X_2=m-r) [/tex]
which after summing, hopefully looks something like what you got...

Then the probabilty of Y< a is:
[tex]P(Y<a) = \sum_m^{a-1} \sum_r P(X_1=r)P(X_2=m-r) [/tex]
note the "a" used in the sum actually represents the integer part of a...

And the probabilty of Y>= a is:
[tex]P(Y>=a) = 1 - P(Y<a) = 1 - \sum_m^{a-1} \sum_r P(X_1=r)P(X_2=m-r)[/tex]

Now we want the proabibility AND case when X1 <=a and Y>=a, so looking at the probabily for X1<= a
[tex] P(X<=a) = \sum_r^a P(X_1=r)[/tex]

Now looking at the expresssion for Y>= a, the X1 cases appear explicitly, so confining the sum to only those cases when X1<=a in the expression should represent the AND case:
[tex](P(Y=X_1+X_2>=a)) \wedge (P(X<=a)) = 1 - \sum_m^{a-1} \sum_r^a P(X_1=r)P(X_2=m-r)[/tex]

what do you think?

 

[bpet]

how do I work out the probability, given an arbitrary positive real a that:

X₁ <= a <= X₁+X₂

Another approach is to draw the 2D grid (X₁,X₂) and highlight the region defined by the above inequality.

 

[lanedance]

nice one... hopefully the sum I gave should in effect count the discrete points within that region divided by (n+1)^2