How do you calculate \(_{0.25}q_{2} \) using the balducci assumption? 2012 UCT paper question 2

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Recall that under Baducci, \( _{1-t}q_{x+t} = (1-t). q_x\). Now, the **crux **is the following:

\( _tp_x\,_{1-t}p_{x+t} = p_x \quad \implies _tq_x= 1 - \frac{p_x}{_{1-t}p_{x+t} } \stackrel{\text{(YOU SHOW)}}= \frac{t.q_x}{1 - (1-t).q_x}\) under Balducci.

The rest should be fairly straightforward from there. Do you agree with my answer of 0.000157578?

I hope I haven't made any typos!

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