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In the dividend-free Black-Scholes model, let $\tau =T-t>0$, $K>0$, $S>0$, $\sigma >0$ andThen the displayed finite sum prices the European option with power put option payoff. Here $\Phi$ is the standard normal distribution function. Indeed, $\log S_T$ conditionally has a normal distribution with mean $\log S+(r-{\sigma }^2/2)\tau$ and variance ${\sigma }^2\tau$. Expand $(K-S_T{)}^n$ by the binomial theorem on $S_T<K$ and apply the lower truncated lognormal moment to each term. Equivalently,This expectation representation proves monotonicity and convexity in $S$. It also gives $P_n(0,t)=K^n{e}^{-r\tau }$ and $P_n(S,t)\to 0$ as $S\to \infty$ by the dominated convergence theorem. For positive $S$ the lognormal distribution smoothing makes the price smooth and strictly positive.

Write $\tau =T-t>0$ and $g_n(s)=(K-s{)}_{+}^n$. The European power put formula isHere $Y>0$ has a lognormal distribution. Since $g_n$ is decreasing and is a convex function, its discounted expected value is also decreasing and convex as a function of $S$. For $S>0$ it is smooth and strictly positive: the normal distribution assigns positive probability to $SY<K$ whatever the current positive spot price. Its endpoints areThe dominated convergence theorem justifies the second limit, because $0\le (K-SY{)}_{+}^n\le K^n$. For a quantitative sketch, the binomial theorem and truncated lognormal moment givewhere $\Phi$ is the standard normal distribution function. This finite sum is useful for drawing the curve; its positive expectation form makes its qualitative shape clearer.

The solid curves use $K=1$, $r=0.05$, $\sigma =0.30$ and $\tau =1$; the dashed curves show their terminal financial payoffs for comparison. Each price remains positive at $S=K$ and tends smoothly towards zero beyond it. At small $S$, discounting gives a price below the undiscounted financial payoff; European exercise is permitted only at expiry.

For the specified row and column, expand the two matrix products and reorder only finite sums:The middle equality uses scalar associativity and distributivity, and the interchange of two finite sums. Together with their equal dimensions, equality of every entry proves $(AB)C=A(BC)$.

For a sum of mutually orthogonal projectors, orthonormality gives $q_i^T{q}_j=0$ for $i\ne j$ and one for $i=j$. Therefore $P_i{P}_j=q_i(q_i^T{q}_j)q_j^T$ equals zero for different indices and $P_i$ for equal indices. Expanding the finite sum yieldsThus every stated partial sum is a projector. Its image of a linear map is the span of the first $k$ orthonormal set of vectors; arbitrary sums of idempotent matrices would not have this conclusion.