# Pure mathematician needs help with probability!

1. Sep 14, 2009

### 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 fY(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 FX1(x) and FY(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') = FX1(x)(1-FY(x)+fY(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 fY/E(x) and FY/E(x),

for then P(E and E') = P(E)P(E'/E) = FX1(x)(1-FY/E(x)+fY/E(x)).

Is this the right way of solving the problem? Is there a way of calculating fY/E(x) and FY/E(x) from fX1(x), FX1(x), fY(x) and FY(x)?

Any advice would be much appeciated,

Thanks

Last edited: Sep 14, 2009
2. Sep 17, 2009

### 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:

$$P(X=m) = \frac{1}{(n+1)}$$
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
$$P(Y=(X_1+X_2)=m) = \sum_r P(X_1=r)P(X_2=m-r)$$
which after summing, hopefully looks something like what you got...

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

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

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

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:
$$(P(Y=X_1+X_2>=a)) \wedge (P(X<=a)) = 1 - \sum_m^{a-1} \sum_r^a P(X_1=r)P(X_2=m-r)$$

what do you think?

Last edited: Sep 17, 2009
3. Sep 25, 2009

### bpet

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

4. Sep 25, 2009

### lanedance

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