Decompose $XX^T = O^T \Lambda O$ with $O$ an $M\times M$ orthogonal matrix and $\Lambda={\rm diag}\,(\lambda_1,\lambda_2,\ldots \lambda_M)$ the diagonal matrix of eigenvalues. Define $w=|v|^{-1} Ov$, then $$v^T XX^T v =|v|^2 \sum_{m=1}^M \lambda_m w_m^2.$$ The matrix $XX^T$ has a Wishart distribution, with independent $O$ and $\Lambda$. It follows that the $w_m$'s are independent Gaussians with mean zero and variance $1/M$. The probability distribution of the $\lambda_m$'s is $$P(\lambda_1,\lambda_2,\ldots\lambda_M)\propto \prod_{m=1}^M e^{-\lambda_m/2}\lambda_m^{(N-M-1)/2}\prod_{i<j}|\lambda_i-\lambda_j|,$$ with $E[\sum_{m}\lambda_m]=NM$.

This gives $$E\left[ {{v^T}X{X^T}{v}} \right]=|v|^2 N.$$ The expectation $E\left[\exp(- {{v^T}X{X^T}{v}}) \right]$ can be evaluated by integration for small $M$, for large $M$ it tends to $e^{-|v|^2 N}$.

Decompose $XX^T = O^T \Lambda O$ with $O$ an $M\times M$ orthogonal matrix and $\Lambda={\rm diag}\,(\lambda_1,\lambda_2,\ldots \lambda_M)$ the diagonal matrix of eigenvalues. Define $w=|v|^{-1} Ov$, then $$v^T XX^T v =|v|^2 \sum_{m=1}^M \lambda_m w_m^2.$$ The matrix $XX^T$ has a Wishart distribution, with independent $O$ and $\Lambda$. It follows that the $w_m$'s are independent Gaussians with mean zero and variance $1/M$. The probability distribution of the $\lambda_m$'s is $$P(\lambda_1,\lambda_2,\ldots\lambda_M)\propto \prod_{m=1}^M e^{-\lambda_m/2}\lambda_m^{(N-M-1)/2}\prod_{i<j}|\lambda_i-\lambda_j|,$$ with $E[\sum_{m}\lambda_m]=NM$.

This gives $$E\left[ {{v^T}X{X^T}{v}} \right]=|v|^2 N.$$ The expectation $E\left[\exp(- {{v^T}X{X^T}{v}}) \right]$ can be evaluated by integration for small $M$, $$E\left[\exp(- {{v^T}X{X^T}{v}}) \right]=\int_0^\infty d\lambda_1\cdots\int_0^\infty d\lambda_M \,P(\lambda_1,\ldots\lambda_M)\prod_{m=1}^M(1+2M|v|^2\lambda_m)^{-1/2},$$ for large $M$ it tends to $e^{-|v|^2 N}$.

Decompose $XX^T = O^T \Lambda O$ with $O$ an $M\times M$ orthogonal matrix and $\Lambda={\rm diag}\,(\lambda_1,\lambda_2,\ldots \lambda_M)$ the diagonal matrix of eigenvalues. Define $w=|v|^{-1} Ov$, then $$v^T XX^T v =|v|^2 \sum_{m=1}^M \lambda_m w_m^2.$$ The matrix $XX^T$ has a Wishart distribution, with independent $O$ and $\Lambda$. It follows that the $w_m$'s are independent Gaussians with mean zero and variance $1/M$. The probability distribution of the $\lambda_m$'s is $$P(\lambda_1,\lambda_2,\ldots\lambda_M)\propto \prod_{m=1}^M e^{-\lambda_m/2}\lambda_m^{(N-M-1)/2}\prod_{i<j}|\lambda_i-\lambda_j|,$$ with $E[\sum_{m}\lambda_m]=NM$.

This gives $$E\left[ {{v^T}X{X^T}{v}} \right]=|v|^2 N.$$ The expectation $E\left[\exp(- {{v^T}X{X^T}{v}}) \right]$ can be evaluated by integration for small $N,M$, for large $N,M$ it tends to $e^{-|v|^2 N}$.

Decompose $XX^T = O^T \Lambda O$ with $O$ an $M\times M$ orthogonal matrix and $\Lambda={\rm diag}\,(\lambda_1,\lambda_2,\ldots \lambda_M)$ the diagonal matrix of eigenvalues. Define $w=|v|^{-1} Ov$, then $$v^T XX^T v =|v|^2 \sum_{m=1}^M \lambda_m w_m^2.$$ The matrix $XX^T$ has a Wishart distribution, with independent $O$ and $\Lambda$. It follows that the $w_m$'s are independent Gaussians with mean zero and variance $1/M$. The probability distribution of the $\lambda_m$'s is $$P(\lambda_1,\lambda_2,\ldots\lambda_M)\propto \prod_{m=1}^M e^{-\lambda_m/2}\lambda_m^{(N-M-1)/2}\prod_{i<j}|\lambda_i-\lambda_j|,$$ with $E[\sum_{m}\lambda_m]=NM$.

This gives $$E\left[ {{v^T}X{X^T}{v}} \right]=|v|^2 N.$$ The expectation $E\left[\exp(- {{v^T}X{X^T}{v}}) \right]$ can be evaluated by integration for small $M$, for large $M$ it tends to $e^{-|v|^2 N}$.