Conditional expectation of variable drawn from uniform distribution with different domains

Questions
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1

I have two variables a and b, both independently drawn from uniform distribution. a is drawn from uniform distribution (m, m+1) while b is drawn from uniform distribution (0,1). Here, m is 0<m<1. The game is not fair and chances of picking a is more. It is theorized simply as -
Condition for picking a is given by : a>= tb. For t as 0<t<1.
Put in words, lower the value of t, more unfairly a is chosen despite b being better.
If we have to compute the expected value of b when a is observed to be chosen then basically we need to compute - E(b|a>=tb). I computed this

(t - m) * Integrate[b, {a, m, t}, {b, 0, (1/t)*a}] / Integrate[1, {a, m, t}, {b, 0, (1/t)*a}] + (1+m-t)*Integrate[b, {a, t, 1 + m}, {b, 0, 1}] / Integrate[1, {a, t, 1 + m}, {b, 0, 1}]

[t-m is the probability that a is between m and t. Similarly, 1+m-t is also the probability].

Am I correct in this approach? Please guide me.

2

I think a direct integration over the conditioned area works:

(* Set parameters *)
m = 1/16;
t = 3/4;

(* Mean by integration *)
bMean = Integrate[b, {a, m, m + 1}, {b, 0, Min[a/t, 1]}]/
        Integrate[1, {a, m, m + 1}, {b, 0, Min[a/t, 1]}] // N
(* 0.410541 *)

This can be checked with simulations:

(* Mean by simulations *)
n = 1000000;
aa = RandomVariate[UniformDistribution[{m, m + 1}], n];
bb = RandomVariate[UniformDistribution[{0, 1}], n];
data = Transpose[{aa, bb}];
data2 = Select[data, #[[1]] >= t  #[[2]] &];
Mean[data2[[All, 2]]]
(* 0.410485 *)

Or one can do it without specific values for m and t:

t =.;
m =.;
bMean = Integrate[b, {a, m, m + 1}, {b, 0, Min[a/t, 1]}, Assumptions -> {0 < m < 1, 0 < t < 1}]/
    Integrate[1, {a, m, m + 1}, {b, 0, Min[a/t, 1]}, Assumptions -> {0 < m < 1, 0 < t < 1}] // PiecewiseExpand // FullSimplify

conditionalexpectation.png

3

Thank you so much!