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For a prime $p$ and an elliptic curve $E/\mathbb{Q}$, we have the exact sequence

$$\displaystyle 0 \rightarrow E(\mathbb{Q})/p E(\mathbb{Q}) \rightarrow S_p(E) \rightarrow \text{Sha}_E[p]\rightarrow 0,$$

where $S_p(E)$ is the $p$-Selmer group of $E$ and $\text{Sha}_E$ is the Tate-Shafarevich group of $E$ and $\text{Sha}_E[p]$ is the $p$-part of it. This gives rise to the equation

$$\displaystyle r_p(S_p(E)) = r(E) + r_p(E(\mathbb{Q}[p]) + r_p (\text{Sha}_E[p]).$$

Here $E(\mathbb{Q})[p]$ is the $p$-torsion subgroup of $E$ (it is always trivial for $p > 11$ (by Mazur's theorem) and usually trivial for $p = 2,3,5,7,11$ (by a density argument)), $r_p$ denotes the $p$-rank of a finite group, and $r(E)$ denotes the Mordell-Weil rank of $E$. On average (by my earlier remark) $r_p(E(\mathbb{Q}[p])$ is zero, and therefore the average rank of $r(E)$ is equal to the average of $r_p(S_p(E)) - r_p(\text{Sha}_E[p])$.

Therefore, if you know the average of $\text{Sha}_p[E]$ for $p = 2,3,5$, then you would know the exact average of the Mordell-Weil rank, since the corresponding average for $p$-Selmer is already known due to work of Bhargava and Shankar. Of course, estimating the average size of Sha is not easier (as far as I know) than estimating the Mordell-Weil rank itself.

However, even if the average comes out to be $1/2$, as expected from conjectures of Goldfeld and Katz-Sarnak, it still doesn't follow that the 'obvious' distribution of half of all curves having rank 0 and the other half having rank 1 would hold. One would need to further control the density of curves having rank $\geq 2$. I am not sure if knowing the size of Sha helps with this question.

For a prime $p$ and an elliptic curve $E/\mathbb{Q}$, we have the exact sequence

$$\displaystyle 0 \rightarrow E(\mathbb{Q})/p E(\mathbb{Q}) \rightarrow S_p(E) \rightarrow \text{Sha}_E[p]\rightarrow 0,$$

where $S_p(E)$ is the $p$-Selmer group of $E$ and $\text{Sha}_E$ is the Tate-Shafarevich group of $E$ and $\text{Sha}_E[p]$ is the $p$-part of it. This gives rise to the equation

$$\displaystyle r_p(S_p(E)) = r(E) + r_p(E(\mathbb{Q}[p]) + r_p (\text{Sha}_E[p]).$$

Here $E(\mathbb{Q})[p]$ is the $p$-torsion subgroup of $E$ (it is always trivial for $p > 11$ (by Mazur's theorem) and usually trivial for $p = 2,3,5,7,11$ (by a density argument)), $r_p$ denotes the $p$-rank of a finite group, and $r(E)$ denotes the Mordell-Weil rank of $E$. On average (by my earlier remark) $r_p(E(\mathbb{Q}[p])$ is zero, and therefore the average rank of $r(E)$ is equal to the average of $r_p(S_p(E)) - r_p(\text{Sha}_E[p])$.

Therefore, if you know the average of $\text{Sha}_p[E]$ for $p = 2,3,5$, then you would know the exact average of the Mordell-Weil rank, since the corresponding average for $p$-Selmer is already known due to work of Bhargava and Shankar. Of course, estimating the average size of Sha is not easier (as far as I know) than estimating the Mordell-Weil rank itself.

However, even if the average comes out to be $1/2$, as expected from conjectures of Goldfeld and Katz-Sarnak, it still doesn't follow that the 'obvious' distribution of half of all curves having rank 0 and the other half having rank 1 would hold. One would need to further control the density of curves having rank $\geq 2$. I am not sure if knowing the size of Sha helps with this question.

I should clarify the connection for those who are unfamiliar with the best known-results on BSD. Essentially, due to important work of many people, it is known that BSD holds for all elliptic curves of rank 0 and 1. Further, it is expected that when ordered by any 'reasonable' height, 50% of elliptic curves have rank 0 and the other half have rank 1. Thus, the latter conjecture implies that 100% of elliptic curves satisfy BSD.

Another remark is that if one assumes the parity conjecture (that is, half of all elliptic curves have odd rank and the other half have even rank, again with respect to a reasonable height), then the Goldfeld/Katz-Sarnak conjecture asserting that the average rank is $1/2$ will be enough to prove that 100% of elliptic curves satisfy BSD. To see this, note that the smallest possible contribution from odd rank curves to the average is $1/2$; which is equivalent to the assertion that 100% of odd rank curves have rank 1 and all larger ranks together occupy 0% of all odd rank curves. This forces 100% of all even rank curves to be rank 0, and then we can conclude that 100% of elliptic curves satisfy BSD.

For a prime $p$ and an elliptic curve $E/\mathbb{Q}$, we have the exact sequence

$$\displaystyle 0 \rightarrow E(\mathbb{Q})/p E(\mathbb{Q}) \rightarrow S_p(E) \rightarrow \text{Sha}_E[p]\rightarrow 0,$$

where $S_p(E)$ is the $p$-Selmer group of $E$ and $\text{Sha}_E$ is the Tate-Shafarevich group of $E$ and $\text{Sha}_E[p]$ is the $p$-part of it. This gives rise to the equation

$$\displaystyle r_p(S_p(E)) = r(E) + r_p(E(\mathbb{Q}[p]) + r_p (\text{Sha}_E[p]).$$

Here $E(\mathbb{Q})[p]$ is the $p$-torsion subgroup of $E$ (it is always empty for $p > 11$ (by Mazur's theorem) and usually empty for $p = 2,3,5,7,11$ (by a density argument)), $r_p$ denotes the $p$-rank of a finite group, and $r(E)$ denotes the Mordell-Weil rank of $E$. On average (by my earlier remark) $r_p(E(\mathbb{Q}[p])$ is zero, and therefore the average rank of $r(E)$ is equal to the average of $r_p(S_p(E)) - r_p(\text{Sha}_E[p])$.

Therefore, if you know the average of $\text{Sha}_p[E]$ for $p = 2,3,5$, then you would know the exact average of the Mordell-Weil rank, since the corresponding average for $p$-Selmer is already known due to work of Bhargava and Shankar. Of course, estimating the average size of Sha is not easier (as far as I know) than estimating the Mordell-Weil rank itself.

However, even if the average comes out to be $1/2$, as expected from conjectures of Goldfeld and Katz-Sarnak, it still doesn't follow that the 'obvious' distribution of half of all curves having rank 0 and the other half having rank 1 would hold. One would need to further control the density of curves having rank $\geq 2$. I am not sure if knowing the size of Sha helps with this question.

For a prime $p$ and an elliptic curve $E/\mathbb{Q}$, we have the exact sequence

$$\displaystyle 0 \rightarrow E(\mathbb{Q})/p E(\mathbb{Q}) \rightarrow S_p(E) \rightarrow \text{Sha}_E[p]\rightarrow 0,$$

where $S_p(E)$ is the $p$-Selmer group of $E$ and $\text{Sha}_E$ is the Tate-Shafarevich group of $E$ and $\text{Sha}_E[p]$ is the $p$-part of it. This gives rise to the equation

$$\displaystyle r_p(S_p(E)) = r(E) + r_p(E(\mathbb{Q}[p]) + r_p (\text{Sha}_E[p]).$$

Here $E(\mathbb{Q})[p]$ is the $p$-torsion subgroup of $E$ (it is always trivial for $p > 11$ (by Mazur's theorem) and usually trivial for $p = 2,3,5,7,11$ (by a density argument)), $r_p$ denotes the $p$-rank of a finite group, and $r(E)$ denotes the Mordell-Weil rank of $E$. On average (by my earlier remark) $r_p(E(\mathbb{Q}[p])$ is zero, and therefore the average rank of $r(E)$ is equal to the average of $r_p(S_p(E)) - r_p(\text{Sha}_E[p])$.

Therefore, if you know the average of $\text{Sha}_p[E]$ for $p = 2,3,5$, then you would know the exact average of the Mordell-Weil rank, since the corresponding average for $p$-Selmer is already known due to work of Bhargava and Shankar. Of course, estimating the average size of Sha is not easier (as far as I know) than estimating the Mordell-Weil rank itself.

However, even if the average comes out to be $1/2$, as expected from conjectures of Goldfeld and Katz-Sarnak, it still doesn't follow that the 'obvious' distribution of half of all curves having rank 0 and the other half having rank 1 would hold. One would need to further control the density of curves having rank $\geq 2$. I am not sure if knowing the size of Sha helps with this question.

For a prime $p$ and an elliptic curve $E/\mathbb{Q}$, we have the exact sequence

$$\displaystyle 0 \rightarrow E(\mathbb{Q})/p E(\mathbb{Q}) \rightarrow S_p(E) \rightarrow \text{Sha}_E[2]\rightarrow 0,$$

where $S_p(E)$ is the $p$-Selmer group of $E$ and $\text{Sha}_E$ is the Tate-Shafarevich group of $E$ and $\text{Sha}_E[p]$ is the $p$-part of it. This gives rise to the equation

$$\displaystyle r_p(S_p(E)) = r(E) + r_p(E(\mathbb{Q}[p]) + r_p (\text{Sha}_E[p]).$$

Here $E(\mathbb{Q})[p]$ is the $p$-torsion subgroup of $E$ (it is always empty for $p > 11$ (by Mazur's theorem) and usually empty for $p = 2,3,5,7,11$ (by a density argument)), $r_p$ denotes the $p$-rank of a finite group, and $r(E)$ denotes the Mordell-Weil rank of $E$. On average (by my earlier remark) $r_p(E(\mathbb{Q}[p])$ is zero, and therefore the average rank of $r(E)$ is equal to the average of $r_p(S_p(E)) - r_p(\text{Sha}_E[p])$.

Therefore, if you know the average of $\text{Sha}_p[E]$ for $p = 2,3,5$, then you would know the exact average of the Mordell-Weil rank, since the corresponding average for $p$-Selmer is already known due to work of Bhargava and Shankar. Of course, estimating the average size of Sha is not easier (as far as I know) than estimating the Mordell-Weil rank itself.

For a prime $p$ and an elliptic curve $E/\mathbb{Q}$, we have the exact sequence

$$\displaystyle 0 \rightarrow E(\mathbb{Q})/p E(\mathbb{Q}) \rightarrow S_p(E) \rightarrow \text{Sha}_E[p]\rightarrow 0,$$

where $S_p(E)$ is the $p$-Selmer group of $E$ and $\text{Sha}_E$ is the Tate-Shafarevich group of $E$ and $\text{Sha}_E[p]$ is the $p$-part of it. This gives rise to the equation

$$\displaystyle r_p(S_p(E)) = r(E) + r_p(E(\mathbb{Q}[p]) + r_p (\text{Sha}_E[p]).$$

Here $E(\mathbb{Q})[p]$ is the $p$-torsion subgroup of $E$ (it is always empty for $p > 11$ (by Mazur's theorem) and usually empty for $p = 2,3,5,7,11$ (by a density argument)), $r_p$ denotes the $p$-rank of a finite group, and $r(E)$ denotes the Mordell-Weil rank of $E$. On average (by my earlier remark) $r_p(E(\mathbb{Q}[p])$ is zero, and therefore the average rank of $r(E)$ is equal to the average of $r_p(S_p(E)) - r_p(\text{Sha}_E[p])$.

Therefore, if you know the average of $\text{Sha}_p[E]$ for $p = 2,3,5$, then you would know the exact average of the Mordell-Weil rank, since the corresponding average for $p$-Selmer is already known due to work of Bhargava and Shankar. Of course, estimating the average size of Sha is not easier (as far as I know) than estimating the Mordell-Weil rank itself.

However, even if the average comes out to be $1/2$, as expected from conjectures of Goldfeld and Katz-Sarnak, it still doesn't follow that the 'obvious' distribution of half of all curves having rank 0 and the other half having rank 1 would hold. One would need to further control the density of curves having rank $\geq 2$. I am not sure if knowing the size of Sha helps with this question.