This Is Why They Don't Do Joint Performance Reviews

I used a,b,c for Rajiv up/flat/down and x,y,z for Karthik.

We have b = 0.05 and

b(y+z) + yc = 0.11

b(x+y) + ya = 0.23

Simply add the two equations, and then use a+b+c=1

b(x+2y+z) + y(a+c) = 0.34

b(1+y) + y(1-b) = 0.34

y = 0.29

Suppose Probability of Rajiv's profit Rp, Loss Rl, No-profit, no-loss Rn Karthik's profit Kp, loss Kl, no-profit no-loss Kn Rn(Kn+Kl)+KnRl=0.11 Rn(Kp+Kn)+KnRp=0.23 Rp+Rl+Rn=Kp+Kl+Kn=1 solving we get Rn+Kn=0.34 Rn is given as 0.05 So Kn is 0.29

I'll use the following notation:

$R_{u}$ is the probability that Rajiv made money on a given day (u for UP)

$R_{f}$ is the probability that Rajiv neither made nor lost any money (f for FLAT)

$R_{d}$ is the probability that Rajiv lost money (d for DOWN)

Same notation for Karthik ( $K_{u}$ , $K_{f}$ , $K_{d}$ )

since one of those three scenarios must happen each day for each trader, we know that:

1) $R_{u}+R_{f}+R_{d} = 1$

2) $K_{u}+K_{f}+K_{d} = 1$

And we are given:

3) $R_{f} = 0.05$

Now we formulate the other probabilities using these variables, assuming that the probabilities of Rajiv and Karthik making money are independent:

4) $R_{f}K_{d}+R_{f}K_{f}+R_{d}K_{f} = 0.11$

5) $R_{f}K_{u}+R_{f}K_{f}+R_{u}K_{f} = 0.23$

We can rewrite 4) as:

$R_{f}K_{d}+R_{f}K_{f}+R_{d}K_{f}+R_{d}K_{d}-R_{d}K_{d} = 0.11$

And factorize:

$(R_{f}+R_{d})(K_{f}+K_{d})-R_{d}K_{d} = 0.11$

Similarly for 5):

$(R_{f}+R_{u})(K_{f}+K_{u})-R_{u}K_{u} = 0.23$

Substituting in for $R_{f}$ and $K_{f}$ from equations 1) and 2) yields:

$(1-R_{u})(1-K_{u})-R_{d}K_{d} = 0.11$

$(1-R_{d})(1-K_{d})-R_{u}K_{u} = 0.23$

Expanding we have:

$1-R_{u}-K_{u}+R_{u}K_{u}-R_{d}K_{d} = 0.11$

$1-R_{d}-K_{d}+R_{d}K_{d}-R_{u}K_{u} = 0.23$

Adding these two equations and simplifying gives:

$1-R_{u}-K_{u}+R_{u}K_{u}-R_{d}K_{d}+1-R_{d}-K_{d}+R_{d}K_{d}-R_{u}K_{u} = 0.11+0.23$

$(1-R_{u}-R_{d})+(1-K_{u}-K_{d}) = 0.34$

$R_{f}+K_{f} = 0.34$

Substituting in the given value for $R_{f}$ (equation 3) leads us to conclude:

$\boxed{K_{f} = 0.29}$