Teaching dimension: Difference between revisions
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Let class <math>{\mathcal C}</math> be sampled as follows: for each <math>f\in{X\choose k}</math>, the concept <math>f</math> is included in <math>{\mathcal C}</math> independently with probability <math>p</math>. The class <math>{\mathcal C}</math> is in fact an Erdős–Rényi random <math>k</math>-uniform hypergraph. | Let class <math>{\mathcal C}</math> be sampled as follows: for each <math>f\in{X\choose k}</math>, the concept <math>f</math> is included in <math>{\mathcal C}</math> independently with probability <math>p</math>. | ||
The class <math>{\mathcal C}</math> is in fact an Erdős–Rényi random <math>k</math>-uniform hypergraph. We denote that <math>m=p{n\choose k}</math>, which is the expected size of <math>{\mathcal C}</math>. | |||
Now we try to upper bound the teaching dimension of <math>{\mathcal C}</math>, i.e. to upper bound the probability of the event <math>\tau({\mathcal C})>t</math> for some threshold <math>t</math>. First, due to union bound, | Now we try to upper bound the teaching dimension of <math>{\mathcal C}</math>, i.e. to upper bound the probability of the event <math>\tau({\mathcal C})>t</math> for some threshold <math>t</math>. First, due to union bound, | ||
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&\le {n\choose k}\Pr[f\in{\mathcal C} \wedge \tau(f)>t]\\ | &\le {n\choose k}\Pr[f\in{\mathcal C} \wedge \tau(f)>t]\\ | ||
&= {n\choose k}\Pr[f\in{\mathcal C}]\cdot\Pr[\tau(f)>t\mid f\in{\mathcal C}]\\ | &= {n\choose k}\Pr[f\in{\mathcal C}]\cdot\Pr[\tau(f)>t\mid f\in{\mathcal C}]\\ | ||
&= | &= m \Pr[\tau(f)>t\mid f\in{\mathcal C}]. | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
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\Pr[X=0] | \Pr[X=0] | ||
\le | \le | ||
\Pr\left[|X-\mathbf{E}[X]|\ge \mathbf{E}[X]\right]\le \frac{\mathbf{Var}[X]}{\mathbf{E}[X]^2}=\frac{\mathbf{E}[X^2]-\mathbf{E}[X]^2}{\mathbf{E}[X]^2}. | \Pr\left[|X-\mathbf{E}[X]|\ge \mathbf{E}[X]\right]\le \frac{\mathbf{Var}[X]}{\mathbf{E}[X]^2}=\frac{\mathbf{E}[X^2]-\mathbf{E}[X]^2}{\mathbf{E}[X]^2}=\frac{\mathbf{E}[X^2]}{\mathbf{E}[X]^2}-1. | ||
</math> | </math> | ||
We then hope that <math>\mathbf{E}[X^2]=\left(1+o\left(m^{-1}\right)\right)\mathbf{E}[X]^2</math>. | |||
:<math> | |||
\begin{align} | |||
\mathbf{E}[X^2] | |||
&= | |||
\mathbf{E}\left[\left(\sum_{S\in{X\choose t}}X_S\right)^2\right]\\ | |||
&= | |||
\mathbf{E}\left[\sum_{S\in{X\choose t}}X_S^2+\sum_{S,T\in{X\choose t}\atop S\neq T}X_SX_T\right]\\ | |||
&= | |||
\mathbf{E}\left[\sum_{S\in{X\choose t}}X_S\right]+\sum_{S,T\in{X\choose t}\atop S\neq T}\mathbf{E}\left[X_SX_T\right]\\ | |||
&= | |||
\mathbf{E}\left[X\right]+\sum_{S,T\in{X\choose t}\atop S\neq T}\Pr\left[\mbox{both }S\mbox{ and }T\mbox{ are teaching sets of }f\right]. | |||
\end{align} | |||
</math> | |||
Then compute <math>\mathbf{E}[X]</math> and the above probability. |
Latest revision as of 10:09, 10 December 2012
Let class [math]\displaystyle{ {\mathcal C} }[/math] be sampled as follows: for each [math]\displaystyle{ f\in{X\choose k} }[/math], the concept [math]\displaystyle{ f }[/math] is included in [math]\displaystyle{ {\mathcal C} }[/math] independently with probability [math]\displaystyle{ p }[/math]. The class [math]\displaystyle{ {\mathcal C} }[/math] is in fact an Erdős–Rényi random [math]\displaystyle{ k }[/math]-uniform hypergraph. We denote that [math]\displaystyle{ m=p{n\choose k} }[/math], which is the expected size of [math]\displaystyle{ {\mathcal C} }[/math].
Now we try to upper bound the teaching dimension of [math]\displaystyle{ {\mathcal C} }[/math], i.e. to upper bound the probability of the event [math]\displaystyle{ \tau({\mathcal C})\gt t }[/math] for some threshold [math]\displaystyle{ t }[/math]. First, due to union bound,
- [math]\displaystyle{ \begin{align} \Pr[\tau({\mathcal C})\gt t] &= \Pr[\exists f\in{\mathcal C}, \tau(f)\gt t]\\ &= \Pr\left[\exists f\in{X\choose k}, f\in{\mathcal C} \wedge \tau(f)\gt t\right]\\ &\le {n\choose k}\Pr[f\in{\mathcal C} \wedge \tau(f)\gt t]\\ &= {n\choose k}\Pr[f\in{\mathcal C}]\cdot\Pr[\tau(f)\gt t\mid f\in{\mathcal C}]\\ &= m \Pr[\tau(f)\gt t\mid f\in{\mathcal C}]. \end{align} }[/math]
Fix a particular [math]\displaystyle{ f\in{\mathcal C} }[/math], the event that [math]\displaystyle{ \tau(f)\gt t }[/math] is equivalent to that [math]\displaystyle{ \forall S\in{X\choose t}, \exists h\in{\mathcal C}\mbox{ that }h\neq f, (h\triangle f)\cap S=\emptyset }[/math]. For every [math]\displaystyle{ S\in{X\choose t} }[/math], let [math]\displaystyle{ X_S }[/math] be the boolean variable indicating that [math]\displaystyle{ S }[/math] is a teaching set of [math]\displaystyle{ f }[/math], i.e. [math]\displaystyle{ \forall h\in{\mathcal C} }[/math] that [math]\displaystyle{ h\neq f }[/math] it holds that [math]\displaystyle{ (h\triangle f)\cap S\neq \emptyset }[/math].
And let
- [math]\displaystyle{ X=\sum_{S\in{X\choose t}}X_S }[/math].
Then
- [math]\displaystyle{ \Pr[\tau(f)\gt t] =\Pr[X=0] }[/math]
which due to the Chebyshev's inequality, is bounded as
- [math]\displaystyle{ \Pr[X=0] \le \Pr\left[|X-\mathbf{E}[X]|\ge \mathbf{E}[X]\right]\le \frac{\mathbf{Var}[X]}{\mathbf{E}[X]^2}=\frac{\mathbf{E}[X^2]-\mathbf{E}[X]^2}{\mathbf{E}[X]^2}=\frac{\mathbf{E}[X^2]}{\mathbf{E}[X]^2}-1. }[/math]
We then hope that [math]\displaystyle{ \mathbf{E}[X^2]=\left(1+o\left(m^{-1}\right)\right)\mathbf{E}[X]^2 }[/math].
- [math]\displaystyle{ \begin{align} \mathbf{E}[X^2] &= \mathbf{E}\left[\left(\sum_{S\in{X\choose t}}X_S\right)^2\right]\\ &= \mathbf{E}\left[\sum_{S\in{X\choose t}}X_S^2+\sum_{S,T\in{X\choose t}\atop S\neq T}X_SX_T\right]\\ &= \mathbf{E}\left[\sum_{S\in{X\choose t}}X_S\right]+\sum_{S,T\in{X\choose t}\atop S\neq T}\mathbf{E}\left[X_SX_T\right]\\ &= \mathbf{E}\left[X\right]+\sum_{S,T\in{X\choose t}\atop S\neq T}\Pr\left[\mbox{both }S\mbox{ and }T\mbox{ are teaching sets of }f\right]. \end{align} }[/math]
Then compute [math]\displaystyle{ \mathbf{E}[X] }[/math] and the above probability.